101. Two rigid bodies are observed in the following ways.
| Body | Observation |
|---|
| X | A line through point \(A\) remains fixed. A point \(B\), \(0.20\,\text{m}\) from this line, has tangential speed \(0.80\,\text{m s}^{-1}\), while point \(C\), \(0.50\,\text{m}\) from it, has tangential speed \(2.00\,\text{m s}^{-1}\). Both remain at their stated perpendicular distances. |
| Y | Three non-collinear marked points undergo the same displacement vector during an interval, and the orientation of the body remains unchanged. |
The most appropriate identification is
ⓐ. X is pure translation and Y is fixed-axis rotation
ⓑ. X is compatible with fixed-axis rotation and Y with pure translation
ⓒ. Both X and Y must be fixed-axis rotations
ⓓ. Neither observation is compatible with rigid-body motion
Correct Answer: X is compatible with fixed-axis rotation and Y with pure translation
Explanation: For body X, the stated line remains fixed and the marked points maintain constant perpendicular distances from it.
Their speed-to-radius ratios are
\[
\frac{0.80}{0.20}=4\,\text{s}^{-1}
\]
and
\[
\frac{2.00}{0.50}=4\,\text{s}^{-1}
\]
Thus both points are compatible with the same angular speed
\[
\omega=4\,\text{rad s}^{-1}
\]
Their velocities are tangential to circles centred on the fixed line, which is the characteristic geometry of fixed-axis rotation.
For body Y, every marked point undergoes the same displacement vector while the orientation remains unchanged.
That is the defining kinematic character of pure translation of a rigid body.
Equal displacements of all points are not the normal signature of non-zero fixed-axis rotation because rotational linear displacement depends on position relative to the axis.
Therefore X represents fixed-axis rotational motion, while Y represents pure translation.
102. A rigid body rotates about the fixed \(z\)-axis. At a certain instant,
\[
\vec r_P=(0.30\hat i+0.40\hat j+0.70\hat k)\,\text{m}
\]
and
\[
\vec v_P=(-2.0\hat i+1.5\hat j)\,\text{m s}^{-1}.
\]
At the same instant another point has
\[
\vec r_Q=(-0.40\hat i+0.30\hat j-0.20\hat k)\,\text{m}.
\]
The velocity of \(Q\) is
ⓐ. \((-1.5\hat i-2.0\hat j)\,\text{m s}^{-1}\)
ⓑ. \((1.5\hat i+2.0\hat j)\,\text{m s}^{-1}\)
ⓒ. \((-2.0\hat i+1.5\hat j)\,\text{m s}^{-1}\)
ⓓ. \((2.0\hat i-1.5\hat j)\,\text{m s}^{-1}\)
Correct Answer: \((-1.5\hat i-2.0\hat j)\,\text{m s}^{-1}\)
Explanation: For rotation about the \(z\)-axis,
\[
\vec v=\vec\omega\times\vec r
\]
Let
\[
\vec\omega=\omega\hat k
\]
\(\textbf{Infer the angular velocity from point P:}\)
\[
\vec v_P
=\omega\hat k\times(0.30\hat i+0.40\hat j+0.70\hat k)
\]
Using
\[
\hat k\times\hat i=\hat j,\qquad
\hat k\times\hat j=-\hat i,
\]
we obtain
\[
\vec v_P
=\omega(-0.40\hat i+0.30\hat j)
\]
Comparison with
\[
-2.0\hat i+1.5\hat j
\]
gives
\[
\omega=5\,\text{rad s}^{-1}
\]
\(\textbf{Apply the same angular velocity to Q:}\)
\[
\vec v_Q
=5\hat k\times(-0.40\hat i+0.30\hat j-0.20\hat k)
\]
The component parallel to the axis contributes zero.
Hence,
\[
\vec v_Q
=5(-0.30\hat i-0.40\hat j)
\]
Therefore,
\[
\vec v_Q=(-1.5\hat i-2.0\hat j)\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\) Option A.
103.
Two marked points \(A\) and \(B\) lie on a straight rigid rod and are \(0.60\,\text{m}\) apart. The rod rotates about a fixed axis perpendicular to the rod and intersecting the segment \(AB\). At a certain instant, \(A\) and \(B\) move in opposite tangential directions with speed magnitudes \(1.0\,\text{m s}^{-1}\) and \(2.0\,\text{m s}^{-1}\), respectively.
The distance of the axis from \(A\) and the angular-speed magnitude are, respectively,
ⓐ. \(0.40\,\text{m}\) and \(2.5\,\text{rad s}^{-1}\)
ⓑ. \(0.30\,\text{m}\) and \(5.0\,\text{rad s}^{-1}\)
ⓒ. \(0.20\,\text{m}\) and \(5.0\,\text{rad s}^{-1}\)
ⓓ. \(0.20\,\text{m}\) and \(10\,\text{rad s}^{-1}\)
Correct Answer: \(0.20\,\text{m}\) and \(5.0\,\text{rad s}^{-1}\)
Explanation: Because the fixed axis intersects the segment \(AB\), the two points are on opposite sides of the axis.
Let their perpendicular distances from the axis be \(r_A\) and \(r_B\).
Then,
\[
r_A+r_B=0.60\,\text{m}. \tag{1}
\]
Both points share the same angular-speed magnitude.
Therefore,
\[
v_A=r_A\omega,\qquad v_B=r_B\omega
\]
Taking the ratio,
\[
\frac{r_A}{r_B}
=\frac{v_A}{v_B}
=\frac{1.0}{2.0}
=\frac12
\]
Hence,
\[
r_B=2r_A
\]
Using Eq. (1),
\[
r_A+2r_A=0.60
\]
Thus,
\[
r_A=0.20\,\text{m},
\qquad
r_B=0.40\,\text{m}
\]
The angular speed is then
\[
\omega=\frac{v_A}{r_A}
=\frac{1.0}{0.20}
=5.0\,\text{rad s}^{-1}
\]
The opposite velocity directions are geometrically consistent with the points lying on opposite sides of the rotation axis.
\(\textbf{Final answer:}\)
\[
\boxed{r_A=0.20\,\text{m},\qquad \omega=5.0\,\text{rad s}^{-1}}
\]
104. A rigid assembly consists of a particle of mass \(2m\) fixed at \(x=-a\) and a particle of mass \(m\) fixed at \(x=+2a\), joined by a light rigid connector. It initially rotates with angular speed \(\omega\) about a fixed \(z\)-axis through \(x=0\). The assembly is then stopped and remounted to rotate with the same angular-speed magnitude about a parallel fixed axis through \(x=a\). The speed of the centre of mass in the first and second arrangements, respectively, is
ⓐ. \(a\omega,\ 0\)
ⓑ. \(2a\omega,\ a\omega\)
ⓒ. \(0,\ 2a\omega\)
ⓓ. \(0,\ a\omega\)
Correct Answer: \(0,\ a\omega\)
Explanation: \(\textbf{Locate the centre of mass:}\)
The centre-of-mass coordinate of the two-mass rigid assembly is
\[
x_{\rm CM}
=\frac{2m(-a)+m(2a)}{3m}
\]
Therefore,
\[
x_{\rm CM}=0
\]
In the first arrangement, the fixed rotation axis passes through this point.
Hence the centre of mass has perpendicular distance
\[
r_{\rm CM}=0
\]
from the axis.
Its speed is therefore
\[
v_{\rm CM,1}=r_{\rm CM}\omega=0
\]
\(\textbf{After remounting:}\)
The physical centre of mass of the assembly is still at the same body-fixed location, but the new parallel axis passes through \(x=a\).
The centre of mass is now a distance
\[
a
\]
from the new axis.
Thus,
\[
v_{\rm CM,2}=a\omega
\]
A centre of mass is not automatically stationary during rotation; it is stationary here only when the chosen fixed axis passes through it.
\(\textbf{Final answer:}\)
\[
\boxed{0,\ a\omega}
\]
105.
A rigid body rotates uniformly about the fixed \(z\)-axis through exactly one-quarter revolution in \(0.50\,\text{s}\). Initially point \(P\) is at \(2a\hat i\), and after the rotation it is at \(-2a\hat j\). Another point \(Q\) is initially at \(-a\hat j\).
The final position vector and instantaneous velocity of \(Q\) are
ⓐ. \(+a\hat i,\ -\pi a\hat j\)
ⓑ. \(-a\hat i,\ +\pi a\hat j\)
ⓒ. \(-a\hat i,\ -\pi a\hat j\)
ⓓ. \(+a\hat i,\ +\pi a\hat j\)
Correct Answer: \(-a\hat i,\ +\pi a\hat j\)
Explanation: Point \(P\) moves from \(+x\) to \(-y\) through a quarter revolution.
Therefore the body's rotation is clockwise as viewed from \(+z\).
Taking anticlockwise as positive,
\[
\Delta\theta=-\frac{\pi}{2}
\]
\(\textbf{Angular velocity:}\)
Because the rotation is uniform,
\[
\omega=\frac{\Delta\theta}{\Delta t}
\]
Thus,
\[
\omega=\frac{-\pi/2}{0.50}=-\pi\,\text{rad s}^{-1},
\]
so
\[
\vec\omega=-\pi\hat k\,\text{rad s}^{-1}
\]
\(\textbf{Rotate point Q through the same quarter-turn:}\)
The initial vector
\[
-a\hat j
\]
rotates clockwise through \(90^\circ\) to
\[
-a\hat i
\]
Hence,
\[
\vec r_{Q,f}=-a\hat i
\]
\(\textbf{Instantaneous velocity at the final position:}\)
\[
\vec v_Q=\vec\omega\times\vec r_{Q,f}
\]
Therefore,
\[
\vec v_Q=(-\pi\hat k)\times(-a\hat i)
\]
Using \(\hat k\times\hat i=\hat j\),
\[
\vec v_Q=+\pi a\hat j
\]
\(\textbf{Final answer:}\)
\[
\boxed{\vec r_{Q,f}=-a\hat i,\qquad
\vec v_Q=+\pi a\hat j}
\]
106. In a rigid body undergoing non-zero rotation about a fixed axis, two distinct particles \(A\) and \(B\) remain stationary. Their coordinates are \(A=(0,0,0)\) and \(B=(0,0,2)\,\text{m}\). Another particle \(C=(0.30,0.40,1.50)\,\text{m}\) has speed \(2.5\,\text{m s}^{-1}\). At the same instant, particle \(D=(-0.40,0.30,-0.60)\,\text{m}\) has speed
ⓐ. \(2.5\,\text{m s}^{-1}\), and the angular speed is \(2.5\,\text{rad s}^{-1}\)
ⓑ. \(5.0\,\text{m s}^{-1}\), and the angular speed is \(2.5\,\text{rad s}^{-1}\)
ⓒ. \(2.0\,\text{m s}^{-1}\), and the angular speed is \(5\,\text{rad s}^{-1}\)
ⓓ. \(2.5\,\text{m s}^{-1}\), and the angular speed is \(5\,\text{rad s}^{-1}\)
Correct Answer: \(2.5\,\text{m s}^{-1}\), and the angular speed is \(5\,\text{rad s}^{-1}\)
Explanation: In non-zero fixed-axis rotation, particles lying on the fixed axis have zero linear speed.
Since both \(A\) and \(B\) are stationary and lie on the same \(z\)-directed line, that line identifies the fixed axis.
Thus the relevant radius of any other point is its perpendicular distance from the \(z\)-axis.
\(\textbf{Use point C to obtain the angular speed:}\)
\[
r_C=\sqrt{(0.30)^2+(0.40)^2}=0.50\,\text{m}
\]
Using
\[
v_C=r_C\omega,
\]
we get
\[
2.5=(0.50)\omega
\]
Therefore,
\[
\omega=5\,\text{rad s}^{-1}
\]
\(\textbf{Radius of point D:}\)
\[
r_D=\sqrt{(-0.40)^2+(0.30)^2}=0.50\,\text{m}
\]
Its \(z\)-coordinate does not alter the perpendicular distance from the axis.
Hence,
\[
v_D=(0.50)(5)=2.5\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\omega=5\,\text{rad s}^{-1},\qquad v_D=2.5\,\text{m s}^{-1}
\]
107. Assertion: Two distinct particles of a rigid body in non-zero fixed-axis rotation can have identical instantaneous linear velocity vectors.
Reason: If their separation vector is parallel to the fixed rotation axis, then \(\vec\omega\times(\vec r_2-\vec r_1)=\vec0\), so their instantaneous velocity vectors are equal.
ⓐ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion
ⓑ. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion
Explanation: For fixed-axis rotation,
\[
\vec v_1=\vec\omega\times\vec r_1
\]
and
\[
\vec v_2=\vec\omega\times\vec r_2.
\]
Subtracting,
\[
\vec v_2-\vec v_1
=\vec\omega\times(\vec r_2-\vec r_1).
\]
If the separation vector \(\vec r_2-\vec r_1\) is parallel to \(\vec\omega\), their cross product is zero.
Therefore,
\[
\vec v_2=\vec v_1.
\]
For example, two points having the same \(x\) and \(y\) coordinates but different \(z\)-coordinates during rotation about the \(z\)-axis have the same perpendicular position from the axis.
They therefore move on equal-radius circles in different planes and can have identical instantaneous velocity vectors.
Equal velocity vectors do not by themselves imply that the entire rigid body is undergoing pure translation.
Thus both statements are true, and the Reason directly establishes the Assertion.
108. A rigid body rotates about a fixed axis parallel to the \(z\)-axis and passing through an unknown point \((x_0,y_0)\) in the \(xy\)-plane. At one instant, point \(P=(2,1)\,\text{m}\) has velocity
\[
\vec v_P=(-2\hat i+4\hat j)\,\text{m s}^{-1},
\]
while point \(Q=(2,4)\,\text{m}\) has velocity
\[
\vec v_Q=(-8\hat i+4\hat j)\,\text{m s}^{-1}.
\]
The axis intersects the \(xy\)-plane at
ⓐ. \((1,0)\,\text{m}\), with \(\omega=4\,\text{rad s}^{-1}\)
ⓑ. \((0,1)\,\text{m}\), with \(\omega=2\,\text{rad s}^{-1}\)
ⓒ. \((0,0)\,\text{m}\), with \(\omega=2\,\text{rad s}^{-1}\)
ⓓ. \((2,0)\,\text{m}\), with \(\omega=4\,\text{rad s}^{-1}\)
Correct Answer: \((0,0)\,\text{m}\), with \(\omega=2\,\text{rad s}^{-1}\)
Explanation: Let the angular velocity be
\[
\vec\omega=\omega\hat k
\]
For a point at \((x,y)\), measured relative to an axis through \((x_0,y_0)\),
\[
\vec r_\perp=(x-x_0)\hat i+(y-y_0)\hat j
\]
Therefore,
\[
\vec v
=\omega\hat k\times\vec r_\perp
=-\omega(y-y_0)\hat i+\omega(x-x_0)\hat j
\]
\(\textbf{Compare the two horizontal velocity components:}\)
For \(P\),
\[
-2=-\omega(1-y_0)
\]
For \(Q\),
\[
-8=-\omega(4-y_0)
\]
Subtracting the first equation from the second,
\[
-6=-3\omega
\]
Hence,
\[
\omega=2\,\text{rad s}^{-1}
\]
\(\textbf{Find the axis coordinates:}\)
Using the common \(+\hat j\) component,
\[
4=\omega(2-x_0)
\]
Thus,
\[
4=2(2-x_0),
\]
which gives
\[
x_0=0
\]
Using the \(P\) \(x\)-component,
\[
-2=-2(1-y_0),
\]
so
\[
y_0=0
\]
Therefore the fixed axis is the \(z\)-axis through the origin.
\(\textbf{Final answer:}\)
\[
(x_0,y_0)=(0,0),\qquad \omega=2\,\text{rad s}^{-1}
\]
109.
The angular velocity of a rigid body varies linearly from \(+4\,\text{rad s}^{-1}\) at \(t=0\) to \(-2\,\text{rad s}^{-1}\) at \(t=3\,\text{s}\). A marked point is \(0.40\,\text{m}\) from the fixed rotation axis.
The net angular displacement of the body and the total distance travelled by the marked point during the \(3\,\text{s}\) interval are
ⓐ. \(5\,\text{rad}\) and \(2.0\,\text{m}\)
ⓑ. \(3\,\text{rad}\) and \(1.2\,\text{m}\)
ⓒ. \(1\,\text{rad}\) and \(2.0\,\text{m}\)
ⓓ. \(3\,\text{rad}\) and \(2.0\,\text{m}\)
Correct Answer: \(3\,\text{rad}\) and \(2.0\,\text{m}\)
Explanation: The area under an \(\omega-t\) graph gives angular displacement.
\(\textbf{Locate the reversal:}\)
The graph falls linearly from \(+4\) to \(-2\) in \(3\,\text{s}\), so its slope is
\[
\alpha=\frac{-2-4}{3}=-2\,\text{rad s}^{-2}
\]
Hence,
\[
\omega=4-2t
\]
The body reverses when
\[
\omega=0,
\]
which occurs at
\[
t=2\,\text{s}
\]
\(\textbf{Signed angular displacement:}\)
From \(0\) to \(2\,\text{s}\), the positive triangular area is
\[
\theta_+=\frac12(2)(4)=4\,\text{rad}
\]
From \(2\) to \(3\,\text{s}\), the negative triangular area has magnitude
\[
|\theta_-|=\frac12(1)(2)=1\,\text{rad}
\]
Therefore,
\[
\Delta\theta_{\rm net}=4-1=3\,\text{rad}
\]
\(\textbf{Total angular distance:}\)
\[
\theta_{\rm dist}=4+1=5\,\text{rad}
\]
The marked point therefore travels
\[
s=r\theta_{\rm dist}
\]
Thus,
\[
s=(0.40)(5)=2.0\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\Delta\theta_{\rm net}=3\,\text{rad},\qquad s=2.0\,\text{m}}
\]
110. At instant \(t_1\), two points \(P\) and \(Q\) of the same rigid body have speed magnitudes \(1.2\,\text{m s}^{-1}\) and \(2.0\,\text{m s}^{-1}\), respectively, while the body rotates anticlockwise. At instant \(t_2\), their speed magnitudes are \(0.9\,\text{m s}^{-1}\) and \(1.5\,\text{m s}^{-1}\), respectively, while the body rotates clockwise about the same fixed axis. Which relation is correct?
ⓐ. \(r_P:r_Q=2:3\) and \(\vec\omega_2=-\frac12\vec\omega_1\)
ⓑ. \(r_P:r_Q=3:5\) and \(\vec\omega_2=-\frac34\vec\omega_1\)
ⓒ. \(r_P:r_Q=3:5\) and \(\vec\omega_2=+\frac34\vec\omega_1\)
ⓓ. \(r_P:r_Q=5:3\) and \(\vec\omega_2=-\frac43\vec\omega_1\)
Correct Answer: \(r_P:r_Q=3:5\) and \(\vec\omega_2=-\frac34\vec\omega_1\)
Explanation: For two points of the same rigid body at any one instant,
\[
v_P=r_P|\omega|,
\qquad
v_Q=r_Q|\omega|
\]
Therefore their speed ratio directly gives their fixed radius ratio.
At \(t_1\),
\[
\frac{r_P}{r_Q}
=\frac{1.2}{2.0}
=\frac35
\]
The second observation provides the same check:
\[
\frac{0.9}{1.5}=\frac35
\]
\(\textbf{Compare angular-speed magnitudes:}\)
Using the same point \(P\),
\[
\frac{|\omega_2|}{|\omega_1|}
=\frac{0.9}{1.2}
=\frac34
\]
The body rotates in opposite senses at the two instants.
Therefore the angular velocity vectors are oppositely directed along the same fixed axis.
Hence,
\[
\vec\omega_2=-\frac34\vec\omega_1
\]
The negative sign represents reversal of rotational sense, while the factor \(3/4\) represents the reduction in angular-speed magnitude.
\(\textbf{Final answer:}\)
\[
\boxed{r_P:r_Q=3:5,\qquad
\vec\omega_2=-\frac34\vec\omega_1}
\]
111. Four proposed data sets are shown for two points \(P\) and \(Q\) of the same rigid body rotating about one fixed axis. In each set, \(r\) is perpendicular distance from the axis, \(s\) is total arc length travelled by the point during the same specified interval, and \(v\) is its speed at the end of that interval.
| Set | \(r_P\) | \(s_P\) | \(v_P\) | \(r_Q\) | \(s_Q\) | \(v_Q\) |
|---|
| I | 0.20 | 0.60 | 1.00 | 0.50 | 1.50 | 2.50 |
| II | 0.20 | 0.60 | 1.00 | 0.50 | 1.00 | 2.50 |
| III | 0.20 | 0.60 | 1.00 | 0.50 | 1.50 | 2.00 |
| IV | 0.20 | 0.40 | 0.80 | 0.50 | 1.50 | 2.00 |
All distances are in metres and speeds in \(\text{m s}^{-1}\). Which set is fully compatible with rigid-body fixed-axis rotation?
ⓐ. Set IV
ⓑ. Set II
ⓒ. Set III
ⓓ. Set I
Correct Answer: Set I
Explanation: Two independent rigid-body conditions must be satisfied.
First, over the same interval, both points undergo the same total angular distance.
Therefore,
\[
\frac{s_P}{r_P}=\frac{s_Q}{r_Q}
\]
Second, at the same final instant they share the same angular-speed magnitude:
\[
\frac{v_P}{r_P}=\frac{v_Q}{r_Q}
\]
\(\textbf{Test Set I:}\)
\[
\frac{0.60}{0.20}=3,
\qquad
\frac{1.50}{0.50}=3
\]
The angular distances agree.
Also,
\[
\frac{1.00}{0.20}=5,
\qquad
\frac{2.50}{0.50}=5
\]
The final angular speeds also agree.
\(\textbf{Why the remaining sets fail:}\)
Set II has unequal \(s/r\), so the two points would require different angular distances.
Set III has equal \(s/r\) but unequal \(v/r\), so its final instantaneous angular velocities would disagree.
Set IV has equal final \(v/r\) but unequal \(s/r\).
Thus only Set I simultaneously satisfies both finite-interval and instantaneous rigid-body conditions.
\(\textbf{Final answer:}\) Set I.
112. Consider the following statements about the rigid-body model used in rotational motion.
Statement I: In an ideal rigid body, the distance between every pair of particles remains fixed.
Statement II: A real solid may be treated approximately as rigid when its deformation is negligible for the motion being studied.
Statement III: If a rotating blade stretches appreciably so that separations between its marked particles change, rigid-body relations cannot automatically be applied to the whole blade as though its geometry were fixed.
Statement IV: The rigid-body approximation means that real inter-particle separations are exactly invariant under every possible applied force.
Which combination is correct?
ⓐ. I and IV only
ⓑ. II, III and IV only
ⓒ. I, II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: The ideal rigid body is a mechanical model in which the separation of every pair of particles remains unchanged.
Statement I therefore gives the defining idealisation correctly.
Real materials can deform, so an actual solid is not perfectly rigid under all possible conditions.
However, when those deformations are negligible compared with the dimensions and motion being analysed, treating the body as rigid is an effective approximation.
That makes Statement II correct.
If a rotating object stretches appreciably, its internal geometry changes.
One can then no longer assume without qualification that all marked points retain fixed relative positions characteristic of one rigid body.
Statement III is consequently correct.
Statement IV incorrectly turns an approximation into an absolute physical claim about real matter under every possible force.
The rigid-body model does not assert that real materials are fundamentally undeformable.
Hence only Statements I, II and III are correct.
113. A composite rigid body consists of a component of mass \(M\) whose own centre of mass is at \(2a\hat i\), and another component of mass \(2M\) whose own centre of mass is at \((-a\hat i+3a\hat j)\). The complete body rotates about the fixed \(z\)-axis through the origin with
\[
\vec\omega=3\hat k\,\text{rad s}^{-1}.
\]
The instantaneous velocity of the centre of mass of the complete body is
ⓐ. \(+6a\hat i\,\text{m s}^{-1}\)
ⓑ. \(+2a\hat j\,\text{m s}^{-1}\)
ⓒ. \(-2a\hat i+6a\hat j\,\text{m s}^{-1}\)
ⓓ. \(-6a\hat i\,\text{m s}^{-1}\)
Correct Answer: \(-6a\hat i\,\text{m s}^{-1}\)
Explanation: The centre of mass of the composite body must first be obtained from the centres of mass of its components.
\(\textbf{Composite centre of mass:}\)
\[
\vec R_{\rm CM}
=\frac{M(2a\hat i)+2M(-a\hat i+3a\hat j)}{3M}
\]
The numerator is
\[
2Ma\hat i-2Ma\hat i+6Ma\hat j
\]
Therefore,
\[
\vec R_{\rm CM}=2a\hat j
\]
Thus the centre of mass lies \(2a\) from the fixed \(z\)-axis.
It behaves kinematically like any other body-fixed point.
\(\textbf{Velocity of the centre of mass:}\)
\[
\vec v_{\rm CM}
=\vec\omega\times\vec R_{\rm CM}
\]
Substituting,
\[
\vec v_{\rm CM}
=3\hat k\times(2a\hat j)
\]
Since
\[
\hat k\times\hat j=-\hat i,
\]
we obtain
\[
\vec v_{\rm CM}=-6a\hat i\,\text{m s}^{-1}
\]
The internal complexity of the body is irrelevant once the location of its overall centre of mass is known.
\(\textbf{Final answer:}\)
\[
\boxed{\vec v_{\rm CM}=-6a\hat i\,\text{m s}^{-1}}
\]
114. At some instant, two marked particles \(P\) and \(Q\) of a rigid body have relative position vector
\[
\vec r_Q-\vec r_P=2a\hat i+a\hat j.
\]
The entire rigid body then rotates through \(90^\circ\) anticlockwise about the fixed \(z\)-axis. Which pair gives the new relative position vector and the separation between \(P\) and \(Q\)?
ⓐ. \(2a\hat i-a\hat j,\ \sqrt5\,a\)
ⓑ. \(-a\hat i+2a\hat j,\ \sqrt5\,a\)
ⓒ. \(a\hat i-2a\hat j,\ 3a\)
ⓓ. \(-2a\hat i-a\hat j,\ \sqrt3\,a\)
Correct Answer: \(-a\hat i+2a\hat j,\ \sqrt5\,a\)
Explanation: The vector joining two particles fixed in a rigid body rotates with the body.
Initially,
\[
\vec r_{QP}=2a\hat i+a\hat j
\]
A \(90^\circ\) anticlockwise rotation about \(+\hat k\) transforms the basis directions according to
\[
\hat i\rightarrow\hat j,
\qquad
\hat j\rightarrow-\hat i
\]
Therefore,
\[
2a\hat i\rightarrow2a\hat j
\]
and
\[
a\hat j\rightarrow-a\hat i
\]
The new relative vector is consequently
\[
\vec r'_{QP}=-a\hat i+2a\hat j
\]
\(\textbf{Rigid-body check:}\)
The initial separation magnitude is
\[
|\vec r_{QP}|=\sqrt{(2a)^2+(a)^2}=\sqrt5\,a
\]
After rotation,
\[
|\vec r'_{QP}|=\sqrt{(-a)^2+(2a)^2}=\sqrt5\,a
\]
Thus the orientation of the separation vector changes while its magnitude remains fixed, exactly as required for an ideal rigid body.
\(\textbf{Final answer:}\)
\[
\boxed{\vec r'_Q-\vec r'_P=-a\hat i+2a\hat j,\qquad |\vec r_Q-\vec r_P|=\sqrt5\,a}
\]
115.
The angular velocity of a rigid body about a fixed axis is \(+4\,\text{rad s}^{-1}\) from \(t=0\) to \(t=2\,\text{s}\), then decreases linearly from \(+4\,\text{rad s}^{-1}\) at \(t=2\,\text{s}\) to \(-4\,\text{rad s}^{-1}\) at \(t=6\,\text{s}\). A marked point lies \(0.25\,\text{m}\) from the axis.
Which pair gives the net angular displacement of the body and the total distance travelled by the marked point from \(t=0\) to \(t=6\,\text{s}\)?
ⓐ. \(8\,\text{rad}\) and \(2\,\text{m}\)
ⓑ. \(0\) and \(4\,\text{m}\)
ⓒ. \(8\,\text{rad}\) and \(4\,\text{m}\)
ⓓ. \(16\,\text{rad}\) and \(4\,\text{m}\)
Correct Answer: \(8\,\text{rad}\) and \(4\,\text{m}\)
Explanation: The area under an \(\omega-t\) graph gives signed angular displacement.
\(\textbf{First constant-angular-speed interval:}\)
\[
\Delta\theta_1=(4)(2)=8\,\text{rad}
\]
From \(t=2\) to \(t=6\,\text{s}\), the angular velocity changes uniformly from \(+4\) to \(-4\,\text{rad s}^{-1}\).
The positive and negative triangular areas over this interval are equal in magnitude, so their signed contributions cancel.
Therefore,
\[
\Delta\theta_{2\to6}=0
\]
Hence the net angular displacement is
\[
\Delta\theta_{\rm net}=8\,\text{rad}
\]
\(\textbf{Angular distance requires unsigned areas:}\)
From \(0\) to \(2\,\text{s}\), the angular distance is \(8\,\text{rad}\).
From \(2\) to \(4\,\text{s}\), the positive triangular area is
\[
\frac12(2)(4)=4\,\text{rad}
\]
From \(4\) to \(6\,\text{s}\), the negative triangular area contributes another \(4\,\text{rad}\) to angular distance.
Thus,
\[
\theta_{\rm distance}=8+4+4=16\,\text{rad}
\]
The marked point therefore travels
\[
s=r\theta_{\rm distance}=(0.25)(16)=4\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\Delta\theta_{\rm net}=8\,\text{rad},\qquad s=4\,\text{m}}
\]
116. The angular position of a rigid body rotating about a fixed axis is
\[
\theta(t)=t^3-6t^2+9t
\]
in radians. At \(t=1\,\text{s}\) and \(t=3\,\text{s}\), respectively, which description of the angular motion is correct?
ⓐ. At both instants \(\omega=0\); \(\alpha\lt 0\) at \(t=1\,\text{s}\) and \(\alpha\gt 0\) at \(t=3\,\text{s}\)
ⓑ. At both instants, \(\omega=0\) and \(\alpha=0\)
ⓒ. At \(t=1\,\text{s}\), \(\omega\gt 0\); at \(t=3\,\text{s}\), \(\omega\lt 0\)
ⓓ. At both instants the body is permanently at rest
Correct Answer: At both instants \(\omega=0\); \(\alpha\lt 0\) at \(t=1\,\text{s}\) and \(\alpha\gt 0\) at \(t=3\,\text{s}\)
Explanation: Angular velocity is the time derivative of angular position:
\[
\omega=\frac{d\theta}{dt}
\]
Therefore,
\[
\omega=3t^2-12t+9
\]
Factoring,
\[
\omega=3(t-1)(t-3)
\]
Hence,
\[
\omega(1)=0,\qquad \omega(3)=0
\]
Angular acceleration is
\[
\alpha=\frac{d\omega}{dt}=6t-12
\]
At \(t=1\,\text{s}\),
\[
\alpha=-6\,\text{rad s}^{-2}
\]
The angular velocity changes from positive to negative there, so \(\theta(t)\) has a local maximum.
At \(t=3\,\text{s}\),
\[
\alpha=+6\,\text{rad s}^{-2}
\]
The angular velocity changes from negative to positive there, so \(\theta(t)\) has a local minimum.
A momentary zero angular velocity therefore need not imply zero angular acceleration or permanent rest.
\(\textbf{Final answer:}\) Option A.
117. At an instant, point P of a rigid body rotating about a fixed axis is \(0.50\,\text{m}\) from the axis. Its tangential speed is \(2.0\,\text{m s}^{-1}\), and the magnitude of its total acceleration is \(10\,\text{m s}^{-2}\). Point Q is \(0.75\,\text{m}\) from the same axis. The magnitude of the angular acceleration and the magnitude of the total acceleration of Q are
ⓐ. \(12\,\text{rad s}^{-2}\) and \(10\,\text{m s}^{-2}\)
ⓑ. \(20\,\text{rad s}^{-2}\) and \(3\sqrt{41}\,\text{m s}^{-2}\)
ⓒ. \(8\,\text{rad s}^{-2}\) and \(6\sqrt5\,\text{m s}^{-2}\)
ⓓ. \(12\,\text{rad s}^{-2}\) and \(15\,\text{m s}^{-2}\)
Correct Answer: \(12\,\text{rad s}^{-2}\) and \(15\,\text{m s}^{-2}\)
Explanation: The total acceleration of an off-axis point has mutually perpendicular tangential and radial components.
\(\textbf{Angular speed from point P:}\)
\[
v_P=\omega r_P
\]
Therefore,
\[
2.0=|\omega|(0.50),
\]
so
\[
|\omega|=4\,\text{rad s}^{-1}
\]
The radial acceleration of P is
\[
a_{rP}=\omega^2r_P
\]
Hence,
\[
a_{rP}=(4)^2(0.50)=8\,\text{m s}^{-2}
\]
\(\textbf{Tangential component at P:}\)
Because \(a_{tP}\perp a_{rP}\),
\[
a_P^2=a_{tP}^2+a_{rP}^2
\]
Thus,
\[
10^2=a_{tP}^2+8^2
\]
Therefore,
\[
a_{tP}=6\,\text{m s}^{-2}
\]
Using
\[
a_{tP}=|\alpha|r_P,
\]
we obtain
\[
|\alpha|=\frac{6}{0.50}=12\,\text{rad s}^{-2}
\]
\(\textbf{Acceleration of point Q:}\)
For \(r_Q=0.75\,\text{m}\),
\[
a_{tQ}=|\alpha|r_Q=(12)(0.75)=9\,\text{m s}^{-2},
\]
and
\[
a_{rQ}=\omega^2r_Q=(4)^2(0.75)=12\,\text{m s}^{-2}
\]
Hence,
\[
a_Q=\sqrt{9^2+12^2}=15\,\text{m s}^{-2}
\]
\(\textbf{Final answer:}\)
\[
\boxed{|\alpha|=12\,\text{rad s}^{-2},\qquad a_Q=15\,\text{m s}^{-2}}
\]
118. A rigid body rotates about a fixed axis parallel to the \(z\)-axis. Two particles \(P\) and \(Q\) have position vectors relative to an origin on that axis:
\[
\vec r_P=(3a\hat i+4a\hat j+2a\hat k),
\qquad
\vec r_Q=(-4a\hat i+3a\hat j-5a\hat k).
\]
At a certain instant, the speed of \(P\) is \(10a\,\text{m s}^{-1}\). The speed of \(Q\) at the same instant is
ⓐ. \(5a\,\text{m s}^{-1}\)
ⓑ. \(10a\,\text{m s}^{-1}\)
ⓒ. \(14a\,\text{m s}^{-1}\)
ⓓ. \(20a\,\text{m s}^{-1}\)
Correct Answer: \(10a\,\text{m s}^{-1}\)
Explanation: For rotation about the \(z\)-axis, the speed depends on perpendicular distance from that axis:
\[
v=r_\perp|\omega|
\]
For point \(P\),
\[
r_P=\sqrt{(3a)^2+(4a)^2}=5a
\]
For point \(Q\),
\[
r_Q=\sqrt{(-4a)^2+(3a)^2}=5a
\]
Their very different \(z\)-coordinates do not enter the perpendicular radius.
Both particles therefore lie at the same radial distance from the fixed axis.
Since all particles of a rigid body share the same instantaneous angular-speed magnitude,
\[
v_P=v_Q
\]
The given speed of \(P\) is
\[
10a\,\text{m s}^{-1}
\]
Therefore,
\[
v_Q=10a\,\text{m s}^{-1}
\]
The two particles may move on different circular paths in different planes, yet their speed magnitudes are equal because those circles have equal radii.
\(\textbf{Final answer:}\)
\[
\boxed{v_Q=10a\,\text{m s}^{-1}}
\]
119. A rigid body first rotates anticlockwise through an angular distance \(5\pi\,\text{rad}\), then clockwise through \(2\pi\,\text{rad}\), and finally anticlockwise through \(\pi\,\text{rad}\). A point \(P\), at radius \(R\), and a point \(Q\), at radius \(3R\), remain fixed in the body. Which statement is correct?
ⓐ. \(Q\) completes three times as many revolutions as \(P\), because its radius is three times larger
ⓑ. Both points have net angular displacement \(4\pi\,\text{rad}\); \(Q\) travels three times the total path length of \(P\)
ⓒ. Both points travel equal total path lengths because they share the same angular motion
ⓓ. The net angular displacement is \(8\pi\,\text{rad}\), because angular distances must always be added without signs
Correct Answer: Both points have net angular displacement \(4\pi\,\text{rad}\); \(Q\) travels three times the total path length of \(P\)
Explanation: Angular displacement is common to every point of a rigid body.
Taking anticlockwise as positive,
\[
\Delta\theta=5\pi-2\pi+\pi=4\pi\,\text{rad}
\]
Thus both \(P\) and \(Q\) undergo the same net angular displacement.
They also undergo the same total angular distance:
\[
\theta_{\rm distance}=5\pi+2\pi+\pi=8\pi\,\text{rad}
\]
However, linear path length depends on radius:
\[
s=r\theta_{\rm distance}
\]
For point \(P\),
\[
s_P=R(8\pi)=8\pi R
\]
For point \(Q\),
\[
s_Q=3R(8\pi)=24\pi R
\]
Hence,
\[
s_Q=3s_P
\]
Their revolution counts are nevertheless identical because revolution count is an angular quantity, not a linear-distance quantity.
\(\textbf{Final answer:}\) Option B.
120. A real rotating component is being tested for whether the rigid-body approximation remains valid. During a short interval, the measured distances between three marked particles change from
\[
PQ=0.40\,\text{m},\quad QR=0.30\,\text{m},\quad RP=0.50\,\text{m}
\]
to
\[
PQ=0.40\,\text{m},\quad QR=0.33\,\text{m},\quad RP=0.50\,\text{m}.
\]
Which conclusion is most appropriate for modelling this motion?
ⓐ. It must still be an ideal rigid body because two of the three distances remain unchanged
ⓑ. It is necessarily undergoing pure translation
ⓒ. The rigid-body approximation fails because the separation \(QR\) changes appreciably
ⓓ. The body must be in fixed-axis rotation because \(PQ\) and \(RP\) are unchanged
Correct Answer: The rigid-body approximation fails because the separation \(QR\) changes appreciably
Explanation: An ideal rigid body is defined by fixed separation between every pair of its constituent particles.
It is not sufficient that most selected distances remain unchanged.
Here,
\[
QR
\]
changes from \(0.30\,\text{m}\) to \(0.33\,\text{m}\).
That is a genuine change in the internal geometry of the marked set.
Consequently the three particles cannot all be treated as points of one ideal rigid body over this interval.
The unchanged values of \(PQ\) and \(RP\) do not repair the failure of the third constraint.
The observation also does not establish pure translation, because translation would require appropriate common displacement behaviour rather than merely selected distance information.
Nor does it establish fixed-axis rotation.
In practical mechanics a slightly deformable solid may still be approximated as rigid when deformation is negligible, but the problem explicitly describes an appreciable change.
Therefore the appropriate idealised conclusion is that the rigid-body approximation fails for this marked configuration.
