401.
A particle moves on a taut light string in a vertical circle. At \(\theta=60^\circ\), measured from the vertically downward direction, its speed is \(5\sqrt2\,\text{m s}^{-1}\) and the string tension is \(27.5\,\text{N}\). At \(\theta=120^\circ\), its speed is \(\sqrt{30}\,\text{m s}^{-1}\) and the tension is \(12.5\,\text{N}\). Take \(g=10\,\text{m s}^{-2}\).
What are the radius, particle mass, and square of its bottom speed?
ⓐ. \(0.5\,\text{m},\;1\,\text{kg},\;60\,\text{m}^2\text{s}^{-2}\)
ⓑ. \(1\,\text{m},\;0.5\,\text{kg},\;60\,\text{m}^2\text{s}^{-2}\)
ⓒ. \(1\,\text{m},\;1\,\text{kg},\;50\,\text{m}^2\text{s}^{-2}\)
ⓓ. \(2\,\text{m},\;0.5\,\text{kg},\;80\,\text{m}^2\text{s}^{-2}\)
Correct Answer: \(1\,\text{m},\;0.5\,\text{kg},\;60\,\text{m}^2\text{s}^{-2}\)
Explanation: The speed difference between two heights follows from energy conservation.
\(\textbf{Infer the radius:}\)
\[
v_{60}^2-v_{120}^2
=
2gR(\cos60^\circ-\cos120^\circ)
\]
Here,
\[
v_{60}^2=50,
\qquad
v_{120}^2=30
\]
Thus
\[
20=2(10)R\left(\frac12+\frac12\right)
\]
Hence
\[
20=20R,
\]
so
\[
R=1\,\text{m}
\]
\(\textbf{Infer the mass from radial dynamics at }60^\circ\textbf{:}\)
\[
T-mg\cos60^\circ=\frac{mv^2}{R}
\]
Therefore
\[
27.5=m\left(50+5\right)
\]
Hence
\[
m=0.5\,\text{kg}
\]
\(\textbf{Recover the bottom speed:}\)
The height at \(60^\circ\) above the bottom is
\[
R(1-\cos60^\circ)=\frac R2
\]
Thus
\[
v_b^2=v_{60}^2+2gR\left(1-\frac12\right)
\]
\[
v_b^2=50+10=60\,\text{m}^2\text{s}^{-2}
\]
\(\textbf{Final answer:}\)
\[
\boxed{R=1\,\text{m},\qquad m=0.5\,\text{kg},\qquad v_b^2=60\,\text{m}^2\text{s}^{-2}}
\]
402. Consider a particle moving upward on a light string in a vertical circle. At some point the string becomes slack for the first time.
Statement I: The string tension is zero at the instant the constraint is first lost.
Statement II: In the absence of an impulse at that instant, the particle's velocity remains continuous.
Statement III: The velocity at the instant of release is tangent to the circle.
Statement IV: After the string becomes slack, one may continue to use the circular radial equation with \(T=0\) at all later positions.
Which of the following is correct?
ⓐ. Statements I and II only
ⓑ. Statements II, III and IV only
ⓒ. Statements I, II and III only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, II and III only
Explanation: A light string can pull but cannot push.
Therefore the limiting condition for first loss of the string constraint is
\[
T=0.
\]
Statement I is correct.
If the string simply becomes slack without an impulsive interaction, the particle's velocity cannot jump discontinuously.
Thus Statement II is correct.
Immediately before loss of constraint, the particle is still moving on the circle, so its velocity is tangential to that circle.
The same velocity continues immediately after release, making Statement III correct.
After the string goes slack, however, the particle is no longer constrained to follow the original circular path.
The radial acceleration need not remain
\[
v^2/R
\]
with respect to the old circle.
Its subsequent motion is determined by the forces that remain, commonly gravity alone.
Statement IV is therefore incorrect.
403. Particle A executes the limiting just-complete vertical circle on a light string. Particle B has the same mass, radius and gravitational field, but its kinetic energy at the lowest point is \(40\%\) greater than that of A. Which set correctly describes B?
ⓐ. \(v_{tB}^2=2gR,\;T_{tB}=mg,\;T_{sB}=4mg\)
ⓑ. \(v_{tB}^2=gR,\;T_{tB}=0,\;T_{sB}=3mg\)
ⓒ. \(v_{tB}^2=4gR,\;T_{tB}=3mg,\;T_{sB}=6mg\)
ⓓ. \(v_{tB}^2=3gR,\;T_{tB}=2mg,\;T_{sB}=5mg\)
Correct Answer: \(v_{tB}^2=3gR,\;T_{tB}=2mg,\;T_{sB}=5mg\)
Explanation: For limiting just-complete motion of A,
\[
v_{bA}^2=5gR
\]
Since kinetic energy is proportional to \(v^2\), increasing the bottom kinetic energy by \(40\%\) gives
\[
v_{bB}^2=1.4(5gR)=7gR
\]
\(\textbf{Top speed of B:}\)
The rise from bottom to top is \(2R\):
\[
v_{tB}^2=v_{bB}^2-4gR
\]
Thus
\[
v_{tB}^2=3gR
\]
At the top,
\[
T_{tB}+mg=\frac{m(3gR)}{R}
\]
Hence
\[
T_{tB}=2mg
\]
\(\textbf{Side-point tension:}\)
The side lies height \(R\) above the bottom:
\[
v_{sB}^2=7gR-2gR=5gR
\]
At the side, gravity has no radial component:
\[
T_{sB}=\frac{mv_{sB}^2}{R}=5mg
\]
\(\textbf{Final answer:}\)
\[
\boxed{v_{tB}^2=3gR,\qquad T_{tB}=2mg,\qquad T_{sB}=5mg}
\]
404. A particle executes a complete vertical circle on a light string. Its tension at the top is \(mg/2\). At what angle \(\theta\), measured from the vertically downward direction, is the tension \(2mg\), and what is the square of the speed there?
ⓐ. \(150^\circ,\;\dfrac52gR\)
ⓑ. \(90^\circ,\;2gR\)
ⓒ. \(120^\circ,\;\dfrac32gR\)
ⓓ. \(120^\circ,\;\dfrac52gR\)
Correct Answer: \(120^\circ,\;\dfrac52gR\)
Explanation: For any complete ideal vertical-circle motion,
\[
T(\theta)=T_t+3mg(1+\cos\theta)
\]
This follows by combining energy conservation with the radial-force equation.
\(\textbf{Use the specified tension:}\)
\[
2mg=\frac12mg+3mg(1+\cos\theta)
\]
Dividing by \(mg\),
\[
2=\frac12+3(1+\cos\theta)
\]
Therefore
\[
\frac32=3(1+\cos\theta)
\]
Hence
\[
1+\cos\theta=\frac12,
\]
so
\[
\cos\theta=-\frac12
\]
Thus
\[
\theta=120^\circ
\]
\(\textbf{Speed from the radial equation:}\)
\[
T-mg\cos\theta=\frac{mv^2}{R}
\]
Therefore
\[
2mg+\frac12mg=\frac{mv^2}{R}
\]
Hence
\[
v^2=\frac52gR
\]
\(\textbf{Final answer:}\)
\[
\boxed{\theta=120^\circ,\qquad v^2=\frac52gR}
\]
405.
A particle moves in a vertical circle on a light string. At the lowest point its speed satisfies \(v_b^2=6gR\). It rises without resistance to the first side point. At that instant a device removes energy \(\frac12mgR\) from the particle without changing the direction of its velocity, and the device then ceases to act.
Which statement correctly describes the subsequent motion?
ⓐ. Immediately after the energy removal \(v^2=2gR\), so the string becomes slack before the top
ⓑ. Immediately after the energy removal \(v^2=4gR\), and the top tension is \(mg\)
ⓒ. Immediately after the energy removal \(v^2=3gR\), and the particle subsequently reaches the top with \(T=0\)
ⓓ. Immediately after the energy removal \(v^2=3gR\), but the particle reaches the top with zero speed
Correct Answer: Immediately after the energy removal \(v^2=3gR\), and the particle subsequently reaches the top with \(T=0\)
Explanation: \(\textbf{Speed on reaching the side before energy removal:}\)
The particle rises by height \(R\) from bottom to side:
\[
v_s^2=v_b^2-2gR.
\]
Thus
\[
v_s^2=6gR-2gR=4gR
\]
Its kinetic energy immediately before the device acts is
\[
K_s=\frac12m(4gR)=2mgR
\]
\(\textbf{Energy removed by the device:}\)
The device extracts
\[
\frac12mgR
\]
Hence the remaining kinetic energy is
\[
K_s'=2mgR-\frac12mgR=\frac32mgR
\]
Therefore
\[
\frac12mv_s'^2=\frac32mgR,
\]
which gives
\[
v_s'^2=3gR.
\]
\(\textbf{Subsequent side-to-top motion:}\)
The particle rises through another height \(R\):
\[
v_t^2=3gR-2gR=gR.
\]
At the top,
\[
T_t+mg=\frac{m(gR)}{R}=mg.
\]
Thus
\[
T_t=0
\]
The energy extraction has placed the particle exactly into the limiting state required to just complete the remaining vertical circle.
\(\textbf{Final answer:}\)
\[
\boxed{v_s'^2=3gR,\qquad v_t^2=gR,\qquad T_t=0}
\]
406.
For a particle moving in a complete vertical circle on a light string, a graph of string tension \(T\) against \(\cos\theta\) is a straight line, where \(\theta\) is measured from the vertically downward direction. The graph passes through \((\cos\theta=-1,\;T=2\,\text{N})\) and \((\cos\theta=+1,\;T=14\,\text{N})\). Take \(g=10\,\text{m s}^{-2}\) and circle radius \(R=0.50\,\text{m}\).
Which set gives the particle mass, bottom speed, and top speed?
ⓐ. \(0.20\,\text{kg},\;\sqrt{30}\,\text{m s}^{-1},\;\sqrt{10}\,\text{m s}^{-1}\)
ⓑ. \(0.40\,\text{kg},\;\sqrt{15}\,\text{m s}^{-1},\;\sqrt5\,\text{m s}^{-1}\)
ⓒ. \(0.20\,\text{kg},\;\sqrt{35}\,\text{m s}^{-1},\;\sqrt{15}\,\text{m s}^{-1}\)
ⓓ. \(0.60\,\text{kg},\;5\,\text{m s}^{-1},\;\sqrt5\,\text{m s}^{-1}\)
Correct Answer: \(0.20\,\text{kg},\;\sqrt{30}\,\text{m s}^{-1},\;\sqrt{10}\,\text{m s}^{-1}\)
Explanation: For ideal vertical-circle motion,
\[
v^2=v_b^2-2gR(1-\cos\theta)
\]
Using
\[
T-mg\cos\theta=\frac{mv^2}{R},
\]
one obtains
\[
T=(T_b-3mg)+3mg\cos\theta
\]
Thus the slope of the \(T\) versus \(\cos\theta\) graph is
\[
3mg
\]
\(\textbf{Extract the mass from the graph:}\)
\[
\text{slope}=\frac{14-2}{1-(-1)}=6\,\text{N}
\]
Therefore
\[
3mg=6
\]
With \(g=10\,\text{m s}^{-2}\),
\[
m=0.20\,\text{kg}
\]
\(\textbf{Bottom speed:}\)
At the bottom,
\[
T_b=14\,\text{N},
\qquad
mg=2\,\text{N}
\]
Hence
\[
14-2=\frac{(0.20)v_b^2}{0.50}
\]
Thus
\[
v_b^2=30,
\]
so
\[
v_b=\sqrt{30}\,\text{m s}^{-1}
\]
\(\textbf{Top speed:}\)
At the top,
\[
T_t+mg=\frac{mv_t^2}{R}
\]
Therefore
\[
2+2=\frac{(0.20)v_t^2}{0.50},
\]
giving
\[
v_t^2=10
\]
\(\textbf{Final answer:}\)
\[
\boxed{m=0.20\,\text{kg},\quad v_b=\sqrt{30}\,\text{m s}^{-1},\quad v_t=\sqrt{10}\,\text{m s}^{-1}}
\]
407. A particle moves upward in a complete vertical circle of radius \(R\) on a light string. Its bottom speed satisfies
\[
v_b^2=7gR.
\]
At the instant when the string makes an angle \(60^\circ\) with the vertically downward direction, which set correctly gives its inward radial acceleration, tangential acceleration, and tension?
ⓐ. \(6g,\;(\sqrt3/2)g\) opposite the motion,\;\(\dfrac{13}{2}mg\)
ⓑ. \(5g,\;(\sqrt3/2)g\) along the motion,\;\(\dfrac{11}{2}mg\)
ⓒ. \(6g,\;g/2\) opposite the motion,\;6mg
ⓓ. \(7g,\;(\sqrt3/2)g\) opposite the motion,\;7mg
Correct Answer: \(6g,\;(\sqrt3/2)g\) opposite the motion,\;\(\dfrac{13}{2}mg\)
Explanation: \(\textbf{Speed at }60^\circ\textbf{:}\)
The height above the bottom is
\[
h=R(1-\cos60^\circ)=\frac R2
\]
Energy conservation gives
\[
v^2=7gR-2g\left(\frac R2\right)
\]
Hence
\[
v^2=6gR
\]
The inward radial acceleration is therefore
\[
a_r=\frac{v^2}{R}=6g
\]
The tangential component of gravity has magnitude
\[
a_t=g\sin60^\circ=\frac{\sqrt3}{2}g
\]
Because the particle is moving upward, this tangential acceleration is opposite to its motion.
\(\textbf{Radial force equation:}\)
\[
T-mg\cos60^\circ=\frac{mv^2}{R}
\]
Thus
\[
T-\frac12mg=6mg
\]
Therefore
\[
T=\frac{13}{2}mg
\]
\(\textbf{Final answer:}\)
\[
\boxed{a_r=6g,\quad a_t=\frac{\sqrt3}{2}g\text{ opposite motion},\quad T=\frac{13}{2}mg}
\]
408.
A particle of mass \(m\) is projected from the lowest point of a vertical circle of radius \(R\) on a light string. From the bottom to the first side point only, a tangential resistive force of constant magnitude \(mg/\pi\) opposes the motion. Beyond the side point the resistance vanishes. Gravity acts throughout.
What is the minimum bottom speed required for the particle to complete the vertical circle, and what is the corresponding bottom tension?
ⓐ. \(\sqrt{5gR},\;6mg\)
ⓑ. \(\sqrt{6gR},\;7mg\)
ⓒ. \(\sqrt{7gR},\;8mg\)
ⓓ. \(2\sqrt{gR},\;5mg\)
Correct Answer: \(\sqrt{6gR},\;7mg\)
Explanation: To complete the remaining motion from the side point to the top in the limiting case, the side speed must satisfy
\[
v_s^2=3gR
\]
Thus the minimum kinetic energy at the side is
\[
K_s=\frac12m(3gR)=\frac32mgR
\]
\(\textbf{Energy lost before reaching the side:}\)
The rise in gravitational potential energy from bottom to side is
\[
\Delta U_g=mgR
\]
The arc length from bottom to side is
\[
s=\frac{\pi R}{2}
\]
Hence the resistive work is
\[
W_r=-\frac{mg}{\pi}\left(\frac{\pi R}{2}\right)
\]
Thus
\[
W_r=-\frac12mgR
\]
If \(K_b\) is the minimum bottom kinetic energy,
\[
K_s=K_b-mgR-\frac12mgR
\]
Therefore
\[
\frac32mgR=K_b-\frac32mgR
\]
Hence
\[
K_b=3mgR
\]
So
\[
\frac12mv_b^2=3mgR,
\]
giving
\[
v_b^2=6gR
\]
At the bottom,
\[
T_b-mg=\frac{m(6gR)}{R}=6mg
\]
Therefore
\[
T_b=7mg
\]
\(\textbf{Final answer:}\)
\[
\boxed{v_{b,\min}=\sqrt{6gR},\qquad T_b=7mg}
\]
409. A particle on a taut light string is moving upward in a vertical circle. At an angle \(60^\circ\) from the vertically downward direction, its speed satisfies
\[
v^2=\frac72gR.
\]
No dissipative force acts. Which set gives the tension at this instant and the minimum additional kinetic energy that would have to be supplied instantaneously at this point, without changing the direction of velocity, so that the particle could just complete the circle?
ⓐ. \(4mg,\;\dfrac14mgR\)
ⓑ. \(\dfrac72mg,\;\dfrac12mgR\)
ⓒ. \(4mg,\;\dfrac12mgR\)
ⓓ. \(\dfrac92mg,\;\dfrac14mgR\)
Correct Answer: \(4mg,\;\dfrac14mgR\)
Explanation: \(\textbf{Current string tension:}\)
At \(60^\circ\),
\[
T-mg\cos60^\circ=\frac{mv^2}{R}
\]
Therefore
\[
T-\frac12mg=\frac72mg
\]
Hence
\[
T=4mg
\]
So the string is comfortably taut at the stated instant.
That does not by itself guarantee that the particle has enough energy to remain taut all the way to the top.
\(\textbf{Minimum speed required at this point:}\)
The height still to be gained before the top is
\[
2R-\frac R2=\frac32R
\]
For the limiting complete motion,
\[
v_t^2=gR
\]
Hence energy conservation from the present point to the top requires
\[
v_{\min}^2=gR+2g\left(\frac32R\right)
\]
Thus
\[
v_{\min}^2=4gR
\]
The required increase in \(v^2\) is
\[
4gR-\frac72gR=\frac12gR
\]
Therefore the additional kinetic energy is
\[
\Delta K=\frac12m\left(\frac12gR\right)
\]
Hence
\[
\Delta K=\frac14mgR
\]
\(\textbf{Final answer:}\)
\[
\boxed{T=4mg,\qquad \Delta K_{\min}=\frac14mgR}
\]
410. A particle moves in a complete vertical circle of radius \(0.50\,\text{m}\) on a light string. Take \(g=10\,\text{m s}^{-2}\). Its tension is \(26\,\text{N}\) at \(\theta=60^\circ\) and \(14\,\text{N}\) at \(\theta=120^\circ\), where \(\theta\) is measured from the vertically downward direction. What are the particle mass, bottom tension, and bottom speed?
ⓐ. \(0.20\,\text{kg},\;20\,\text{N},\;5\,\text{m s}^{-1}\)
ⓑ. \(0.40\,\text{kg},\;32\,\text{N},\;\sqrt{35}\,\text{m s}^{-1}\)
ⓒ. \(0.40\,\text{kg},\;28\,\text{N},\;\sqrt{30}\,\text{m s}^{-1}\)
ⓓ. \(0.60\,\text{kg},\;38\,\text{N},\;\sqrt{25}\,\text{m s}^{-1}\)
Correct Answer: \(0.40\,\text{kg},\;32\,\text{N},\;\sqrt{35}\,\text{m s}^{-1}\)
Explanation: For an ideal vertical circle,
\[
T(\theta)=T_b-3mg(1-\cos\theta)
\]
\(\textbf{Use the two measured tensions:}\)
Subtracting,
\[
T_{60}-T_{120}
=
3mg(\cos60^\circ-\cos120^\circ)
\]
Thus
\[
26-14
=
3mg\left(\frac12+\frac12\right)
\]
Hence
\[
12=3mg
\]
Therefore
\[
mg=4\,\text{N},
\]
so
\[
m=0.40\,\text{kg}
\]
\(\textbf{Recover the bottom tension:}\)
At \(60^\circ\),
\[
26=T_b-3(4)\left(1-\frac12\right)
\]
Therefore
\[
26=T_b-6,
\]
giving
\[
T_b=32\,\text{N}
\]
\(\textbf{Bottom speed:}\)
\[
T_b-mg=\frac{mv_b^2}{R}
\]
Hence
\[
32-4=\frac{(0.40)v_b^2}{0.50}
\]
Thus
\[
28=0.8v_b^2
\]
Therefore
\[
v_b^2=35
\]
\(\textbf{Final answer:}\)
\[
\boxed{m=0.40\,\text{kg},\quad T_b=32\,\text{N},\quad v_b=\sqrt{35}\,\text{m s}^{-1}}
\]
411. A particle executes a complete vertical circle on a light string with
\[
v_b^2=8gR.
\]
The following table lists proposed values at four angular positions.
| Row | \(\theta\) | Proposed \(v^2\) | Proposed tension |
|---|
| P | \(0^\circ\) | \(8gR\) | \(9mg\) |
| Q | \(60^\circ\) | \(7gR\) | \(\dfrac{15}{2}mg\) |
| R | \(120^\circ\) | \(5gR\) | \(\dfrac92mg\) |
| S | \(180^\circ\) | \(4gR\) | \(2mg\) |
Which row is inconsistent with the same ideal motion?
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: At angle \(\theta\),
\[
v^2=v_b^2-2gR(1-\cos\theta).
\]
With
\[
v_b^2=8gR,
\]
this becomes
\[
v^2=gR(6+2\cos\theta).
\]
The radial equation gives
\[
T-mg\cos\theta=\frac{mv^2}{R}.
\]
Thus
\[
\frac{T}{mg}=6+3\cos\theta.
\]
At the bottom,
\[
T=9mg,
\]
so P is correct.
At \(60^\circ\),
\[
T=mg\left(6+\frac32\right)=\frac{15}{2}mg,
\]
so Q is correct.
At \(120^\circ\),
\[
T=mg\left(6-\frac32\right)=\frac92mg,
\]
so R is correct.
At the top,
\[
T=mg(6-3)=3mg.
\]
The speed value
\[
v_t^2=4gR
\]
is correct, but the proposed tension \(2mg\) is not.
Hence row S is inconsistent.
412. A particle on a light string executes the limiting just-complete vertical circle of radius \(R\). Let \(a_b,a_s,a_t\) be the magnitudes of its total acceleration at the bottom, side and top, respectively. Which set is correct?
ⓐ. \(a_b=5g,\;a_s=3g,\;a_t=g\)
ⓑ. \(a_b=6g,\;a_s=3g,\;a_t=0\)
ⓒ. \(a_b=5g,\;a_s=\sqrt{10}\,g,\;a_t=g\)
ⓓ. \(a_b=6g,\;a_s=\sqrt{10}\,g,\;a_t=g\)
Correct Answer: \(a_b=5g,\;a_s=\sqrt{10}\,g,\;a_t=g\)
Explanation: For limiting just-complete motion,
\[
v_b^2=5gR,\qquad v_s^2=3gR,\qquad v_t^2=gR
\]
\(\textbf{Bottom:}\)
The tangential component of gravity is zero.
Thus the acceleration is purely radial:
\[
a_b=\frac{v_b^2}{R}=5g
\]
\(\textbf{Side point:}\)
The radial acceleration is
\[
a_{r,s}=3g
\]
Gravity is entirely tangential at the side, giving
\[
a_{t,s}=g
\]
The radial and tangential components are perpendicular, so
\[
a_s=\sqrt{(3g)^2+g^2}
\]
Hence
\[
a_s=\sqrt{10}\,g
\]
\(\textbf{Top:}\)
Again the tangential component of gravity is zero.
The radial acceleration is
\[
a_t=\frac{gR}{R}=g
\]
Thus
\[
5g\gt \sqrt{10}\,g\gt g
\]
\(\textbf{Final answer:}\)
\[
\boxed{a_b=5g,\qquad a_s=\sqrt{10}\,g,\qquad a_t=g}
\]
413. A particle just completes a vertical circle of radius \(R\) on a light string in a gravitational field of magnitude \(g\). Without changing its bottom speed or the radius, the same experiment is then performed in a gravitational field of magnitude \(g/2\). Let \(W'\) denote the particle's weight in the new field. Which statement is correct for the second experiment?
ⓐ. The particle fails to complete the circle because the required centripetal force is smaller
ⓑ. It just completes the circle again with zero top tension
ⓒ. It completes the circle with top tension \(5W'\)
ⓓ. It completes the circle with top tension \(2W'\)
Correct Answer: It completes the circle with top tension \(5W'\)
Explanation: In the original field, limiting completion requires
\[
v_b^2=5gR.
\]
The same bottom speed is retained in the new field.
Let the new gravitational acceleration be
\[
g'=\frac g2.
\]
\(\textbf{Top speed in the new field:}\)
Energy conservation gives
\[
v_t^2=v_b^2-4g'R.
\]
Therefore
\[
v_t^2=5gR-4\left(\frac g2\right)R
\]
Hence
\[
v_t^2=3gR
\]
The new weight is
\[
W'=mg'=\frac12mg
\]
At the top,
\[
T_t+W'=\frac{mv_t^2}{R}.
\]
Thus
\[
T_t+\frac12mg=3mg
\]
Therefore
\[
T_t=\frac52mg
\]
Since
\[
W'=\frac12mg,
\]
\[
T_t=5W'.
\]
The reduced gravity makes the original bottom speed more than sufficient for completion.
\(\textbf{Final answer:}\)
\[
T_t=5W'.
\]
414.
For a \(0.50\,\text{kg}\) particle undergoing limiting just-complete vertical circular motion on a light string, a graph of \(v^2\) against vertical height \(h\) above the lowest point is a straight line. The graph passes through \((h=0,\;v^2=70\,\text{m}^2\text{s}^{-2})\) and \((h=2\,\text{m},\;v^2=30\,\text{m}^2\text{s}^{-2})\).
What are \(g\), the circle radius, and the tension at the lowest point?
ⓐ. \(10\,\text{m s}^{-2},\;1.4\,\text{m},\;30\,\text{N}\)
ⓑ. \(10\,\text{m s}^{-2},\;2.0\,\text{m},\;22.5\,\text{N}\)
ⓒ. \(5\,\text{m s}^{-2},\;2.8\,\text{m},\;15\,\text{N}\)
ⓓ. \(20\,\text{m s}^{-2},\;0.70\,\text{m},\;35\,\text{N}\)
Correct Answer: \(10\,\text{m s}^{-2},\;1.4\,\text{m},\;30\,\text{N}\)
Explanation: For ideal motion,
\[
v^2=v_b^2-2gh
\]
Therefore the slope of the \(v^2-h\) graph is
\[
-2g
\]
\(\textbf{Extract }g\textbf{ from the graph:}\)
\[
\text{slope}=\frac{30-70}{2-0}=-20
\]
Hence
\[
-2g=-20,
\]
so
\[
g=10\,\text{m s}^{-2}
\]
The graph equation is therefore
\[
v^2=70-20h
\]
\(\textbf{Use the limiting top condition:}\)
At the top,
\[
h=2R
\]
and
\[
v_t^2=gR=10R
\]
Therefore
\[
10R=70-20(2R)
\]
Thus
\[
10R=70-40R
\]
Hence
\[
R=1.4\,\text{m}
\]
\(\textbf{Bottom tension:}\)
\[
T_b-mg=\frac{mv_b^2}{R}
\]
Substituting,
\[
T_b-5=\frac{(0.50)(70)}{1.4}
\]
Therefore
\[
T_b-5=25,
\]
so
\[
T_b=30\,\text{N}
\]
\(\textbf{Final answer:}\)
\[
\boxed{g=10\,\text{m s}^{-2},\qquad R=1.4\,\text{m},\qquad T_b=30\,\text{N}}
\]
415. Consider a particle moving from the lowest point to the highest point of a vertical circle on a taut light string, with gravity as the only force doing work.
Statement I: The speed decreases continuously during the upward half-circle.
Statement II: The string tension decreases continuously during the upward half-circle.
Statement III: The tension does zero work throughout the motion.
Statement IV: The instantaneous power delivered by gravity is negative at every point from the bottom through the top, including both endpoints.
Which of the following is correct?
ⓐ. Statements I and II only
ⓑ. Statements II, III and IV only
ⓒ. Statements I, II and III only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, II and III only
Explanation: As the particle rises, its gravitational potential energy increases.
Because tension does no work, this increase comes from kinetic energy.
Hence the speed decreases continuously, so Statement I is correct.
For ideal string motion,
\[
T=T_b-3mg(1-\cos\theta).
\]
During the upward half-circle, \(\theta\) increases from \(0\) to \(\pi\), so \(\cos\theta\) decreases continuously.
Therefore the tension also decreases continuously, making Statement II correct.
Tension is always radial while the instantaneous displacement is tangential.
Thus
\[
dW_T=0,
\]
so Statement III is correct.
Gravity generally delivers negative power during the upward motion because the velocity has an upward component.
At the exact bottom and exact top, however, the velocity is horizontal while gravity is vertical.
Therefore
\[
P_g=0
\]
at both endpoints.
Statement IV is therefore incorrect.
416. A particle on a taut light string is moving upward in a vertical circle. At the instant when the string makes an angle \(120^\circ\) with the vertically downward direction,
\[
v^2=2gR.
\]
No non-conservative force acts. Which statement is correct?
ⓐ. The tension is \(mg/2\), and the string becomes slack before reaching the top
ⓑ. The tension is \(3mg/2\), and the particle subsequently just completes the circle
ⓒ. The tension is \(2mg\), and the top tension becomes \(mg\)
ⓓ. The tension is zero at this point, so the particle leaves the circle immediately
Correct Answer: The tension is \(3mg/2\), and the particle subsequently just completes the circle
Explanation: At
\[
\theta=120^\circ,
\qquad
\cos\theta=-\frac12
\]
\(\textbf{Local radial condition:}\)
\[
T-mg\cos\theta=\frac{mv^2}{R}.
\]
Thus
\[
T+\frac12mg=2mg
\]
Therefore
\[
T=\frac32mg
\]
The string is clearly taut at this point.
\(\textbf{Energy remaining to the top:}\)
At \(120^\circ\), the height above the bottom is
\[
R\left(1+\frac12\right)=\frac32R
\]
The top lies at height \(2R\), so the remaining rise is
\[
\frac12R
\]
Hence
\[
v_t^2=2gR-2g\left(\frac12R\right)
\]
Thus
\[
v_t^2=gR.
\]
At the top,
\[
T_t+mg=\frac{m(gR)}{R}=mg.
\]
Therefore
\[
T_t=0
\]
The particle reaches exactly the limiting just-complete condition.
\(\textbf{Final answer:}\)
\[
T=\frac32mg,\qquad T_t=0
\]
417. A particle is projected from the lowest point of a vertical circle on a light string. During its motion from bottom to top, a non-conservative interaction removes a total mechanical energy \(mgR\). The particle nevertheless reaches the top with the string just taut. Assume the circular constraint remains valid until the top. What are the required bottom speed and the bottom tension?
ⓐ. \(\sqrt{5gR},\;6mg\)
ⓑ. \(\sqrt{6gR},\;7mg\)
ⓒ. \(\sqrt{7gR},\;8mg\)
ⓓ. \(2\sqrt{2gR},\;9mg\)
Correct Answer: \(\sqrt{7gR},\;8mg\)
Explanation: At the top of the limiting motion,
\[
T_t=0
\]
Therefore
\[
mg=\frac{mv_t^2}{R}
\]
Hence
\[
v_t^2=gR
\]
Thus the top kinetic energy is
\[
K_t=\frac12mgR
\]
\(\textbf{Bottom-to-top energy balance with loss:}\)
The increase in gravitational potential energy is
\[
\Delta U=2mgR
\]
The stated non-conservative work is
\[
W_{\text{nc}}=-mgR
\]
Using
\[
K_t+U_t=K_b+U_b+W_{\text{nc}},
\]
with \(U_b=0\),
\[
\frac12mgR+2mgR=K_b-mgR
\]
Therefore
\[
K_b=\frac72mgR
\]
Hence
\[
\frac12mv_b^2=\frac72mgR,
\]
so
\[
v_b^2=7gR
\]
\(\textbf{Bottom tension:}\)
\[
T_b-mg=\frac{m(7gR)}{R}=7mg
\]
Thus
\[
T_b=8mg
\]
\(\textbf{Final answer:}\)
\[
\boxed{v_b=\sqrt{7gR},\qquad T_b=8mg}
\]
418. Particle A of mass \(m\) and Particle B of mass \(2m\) move separately on light strings of the same radius \(R\) in the same gravitational field. Each particle is given the same kinetic energy
\[
5mgR
\]
at the lowest point, where \(m\) denotes the mass of Particle A. Which statement is correct?
ⓐ. A just completes the circle, whereas B fails before the top
ⓑ. Both just complete the circle
ⓒ. A completes with top tension \(5mg\), whereas B just completes with zero top tension
ⓓ. A and B have the same top tension because their bottom kinetic energies are equal
Correct Answer: A completes with top tension \(5mg\), whereas B just completes with zero top tension
Explanation: \(\textbf{Particle A:}\)
Its bottom kinetic energy is
\[
\frac12m v_{bA}^2=5mgR
\]
Thus
\[
v_{bA}^2=10gR
\]
At the top,
\[
v_{tA}^2=10gR-4gR=6gR.
\]
Hence
\[
T_{tA}+mg=6mg
\]
Therefore
\[
T_{tA}=5mg
\]
\(\textbf{Particle B:}\)
Its mass is \(2m\), but its bottom kinetic energy is still \(5mgR\):
\[
\frac12(2m)v_{bB}^2=5mgR
\]
Thus
\[
v_{bB}^2=5gR
\]
This is exactly the limiting bottom-speed condition.
Therefore
\[
v_{tB}^2=gR.
\]
At the top,
\[
T_{tB}+2mg=\frac{(2m)(gR)}{R}=2mg
\]
Hence
\[
T_{tB}=0
\]
Equal kinetic energy does not imply equal speed when the masses differ.
\(\textbf{Final answer:}\)
\[
T_{tA}=5mg,\qquad \text{B just completes with }T_{tB}=0.
\]
419. Two trials use the same particle, circle radius and gravitational field. At \(\theta=120^\circ\), Trial P has
\[
v_P^2=2gR,
\]
so that it subsequently just completes the circle. At the same angular position in Trial Q, the measured string tension is \(3mg\). What are the bottom speed squared and top tension in Trial Q?
ⓐ. \(v_b^2=6gR,\;T_t=mg\)
ⓑ. \(v_b^2=\dfrac{13}{2}gR,\;T_t=\dfrac32mg\)
ⓒ. \(v_b^2=7gR,\;T_t=2mg\)
ⓓ. \(v_b^2=\dfrac{11}{2}gR,\;T_t=\dfrac12mg\)
Correct Answer: \(v_b^2=\dfrac{13}{2}gR,\;T_t=\dfrac32mg\)
Explanation: At
\[
\theta=120^\circ,
\qquad
\cos\theta=-\frac12
\]
\(\textbf{Infer Trial Q speed from the measured tension:}\)
\[
T-mg\cos\theta=\frac{mv^2}{R}
\]
Thus
\[
3mg+\frac12mg=\frac{mv_Q^2}{R}
\]
Therefore
\[
v_Q^2=\frac72gR
\]
\(\textbf{Recover the bottom speed:}\)
The height at \(120^\circ\) is
\[
h=\frac32R
\]
Hence
\[
v_Q^2=v_b^2-2g\left(\frac32R\right)
\]
Therefore
\[
\frac72gR=v_b^2-3gR
\]
Thus
\[
v_b^2=\frac{13}{2}gR
\]
\(\textbf{Top tension:}\)
The top is a further height \(R/2\) above the \(120^\circ\) point:
\[
v_t^2=\frac72gR-gR
\]
Hence
\[
v_t^2=\frac52gR
\]
At the top,
\[
T_t+mg=\frac{m}{R}\left(\frac52gR\right)
\]
Thus
\[
T_t=\frac32mg
\]
\(\textbf{Final answer:}\)
\[
\boxed{v_b^2=\frac{13}{2}gR,\qquad T_t=\frac32mg}
\]
420. Particles A and B have equal masses and move separately on light strings in the same gravitational field. Particle A has circle radius \(R\), while Particle B has radius \(2R\). Both are projected from the bottom with the same speed satisfying
\[
v_b^2=6gR.
\]
Which statement is correct?
ⓐ. Both complete their circles with the same top tension
ⓑ. A just completes, while B loses tautness at the side point
ⓒ. A has top tension \(mg\), while B loses tautness when \(\cos\theta=-1/3\)
ⓓ. A has top tension \(2mg\), while B also completes with zero top tension
Correct Answer: A has top tension \(mg\), while B loses tautness when \(\cos\theta=-1/3\)
Explanation: \(\textbf{Particle A:}\)
For radius \(R\),
\[
v_{bA}^2=6gR.
\]
At the top,
\[
v_{tA}^2=6gR-4gR=2gR.
\]
Hence
\[
T_{tA}+mg=2mg,
\]
so
\[
T_{tA}=mg.
\]
Thus A completes the circle.
\(\textbf{Particle B:}\)
Its radius is
\[
R_B=2R.
\]
The same bottom speed can be written as
\[
v_{bB}^2=3gR_B.
\]
Let
\[
\alpha=\frac{v_{bB}^2}{gR_B}=3.
\]
For ideal circular continuation,
\[
\frac{T}{mg}=\alpha-2+3\cos\theta
\]
The string first becomes slack when
\[
T=0.
\]
Therefore
\[
0=3-2+3\cos\theta
\]
Hence
\[
\cos\theta=-\frac13
\]
So B cannot remain constrained up to the top.
\(\textbf{Final answer:}\)
\[
\boxed{T_{tA}=mg,\qquad \cos\theta_{\text{slack},B}=-\frac13}
\]
