Two masses, $ m_1 $ and $ m_2 $, are suspended on either side of a pulley with a radius $ R $, as shown. The heavier mass, $ m_2 $, is initially held at rest above the ground by a distance $ h $ before being released. An student measures that it takes an amount of time $ t $ for the heavier mass to hit the ground after being released.(a) The pulley is an odd shape, so the student would like to measure its moment of inertia. Derive an algebraic expression for the moment of inertia of the pulley in terms of only quantities measured or known by the student ($ m_1, m_2, R, h, t $ and physical constants). , (b) How many radians has the pulley rotated from the time it was released to the time $ m_2 $ hit the ground? Express your answer in terms of $ m_1, m_2, R, h, t$ and physical constants. :

$\frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1})$ $\frac{h}{R}$

Two masses, $ m_1 $ and $ m_2 $, are suspended on either side of a pulley with a radius $ R $, as shown. The heavier mass, $ m_2 $, is initially held at rest above the ground by a distance $ h $ before being released. An experimenting elephant measures that it takes an amount of time $ t $ for the heavier mass to hit the ground after being released.\n(a) The pulley is an odd shape, and the elephant would like to measure its moment of inertia. Derive an algebraic expression for the moment of inertia of the pulley in terms of only quantities measured or known by the elephant ($ m_1, m_2, R, h, t $ and physical constants).\n(b) How many radians has the pulley rotated from the time it was released to the time $ m_2 $ hit the ground? Express your answer in terms of $ m_1, m_2, R, h $ and physical constants. • Nerd Notes

AP Physics

Unit 6 - Rotational Motion

FRQ

Mathematical

Advanced

Part (a): Moment of Inertia

Step	Derivation/Formula	Reasoning
1	$$ T_{1} – m_{1}g = m_{1}a $$	This is Newton’s second law for mass $m_{1}$ moving upward.
2	$$ m_{2}g – T_{2} = m_{2}a $$	This is Newton’s second law for mass $m_{2}$ moving downward.
3	$$ T_{1} = m_{1}g + m_{1}a \quad \text{and} \quad T_{2} = m_{2}g – m_{2}a $$	Rearrange the equations to solve for the tensions in the string.
4	$$ (T_{2} – T_{1})R = I\left(\frac{a}{R}\right) $$	This relates the net torque on the pulley to its moment of inertia $I$ using the no‐slip condition $\alpha = \frac{a}{R}$.
5	$$ T_{2} – T_{1} = \frac{I\,a}{R^{2}} $$	Simplify the torque equation by dividing both sides by $R$.
6	$$ (m_{2}g – m_{2}a) – (m_{1}g + m_{1}a) = $$$$(m_{2}-m_{1})g – (m_{2}+m_{1})a =$$$$\frac{I\,a}{R^{2}} $$	Substitute the expressions for $T_{1}$ and $T_{2}$ into the torque equation.
7	$$ I = \frac{R^{2}}{a}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})a\Bigr] $$	Rearrange the equation to solve for the moment of inertia $I$.
8	$$ h = \frac{1}{2}at^{2} $$	Use the kinematics relation for the heavy mass $m_{2}$ falling a distance $h$ from rest.
9	$$ a = \frac{2h}{t^{2}} $$	Solve for the acceleration $a$ from the kinematics equation.
10	$$ I = \frac{R^{2}}{\frac{2h}{t^{2}}}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})\frac{2h}{t^{2}}\Bigr] $$	Substitute $a = \frac{2h}{t^{2}}$ into the expression for $I$.
11	$$ I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1}) $$	Simplify the expression to obtain $I$ solely in terms of $m_{1}, m_{2}, R, h, t$ and $g$.
12	$$ \boxed{I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1})} $$	This is the final algebraic expression for the pulley’s moment of inertia.

Part (b): Pulley Rotation

Step	Derivation/Formula	Reasoning
1	$$ \Delta x = R\theta $$	This equation relates the linear displacement $\Delta x$ to the angular displacement $\theta$ of the pulley.
2	$$ h = R\theta $$	Since the heavy mass $m_{2}$ falls a distance $h$, the length of the unwound rope is $h$, which equals $R\theta$.
3	$$ \theta = \frac{h}{R} $$	Solve for the angular displacement $\theta$ of the pulley.
4	$$ \boxed{\theta = \frac{h}{R}} $$	This is the final expression for the total rotation of the pulley in radians.

Need Help? Ask Phy To Explain

A Major Upgrade To Phy Is Coming Soon — Stay Tuned

Just Drag and Drop!

Quick Actions ?

NEW UBQ QUIZ LAB

100s of AP aligned questions and quizzes to help you get a 5 even faster. Full Mock exams with Auto Grading and Adaptive explanations. Try out Nerd Notes', state of the art, quiz platform.

Topics in this question

Dynamic Torque, Rotational Motion, Torque

Need 1 to 1 Help? Book a trial lesson here.

We'll help clarify entire units in one hour or less — guaranteed.

Need to Learn AP Physics From Scratch Quickly? We have Program for that.

A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.

Go Pro to remove ads + unlimited access to our AI learning tools.

Related Questions

5 More similar questions to the one you just solved

Beginner

Conceptual

MCQ

A point P is at a distance $ R $ from the axis of rotation of a rigid body whose angular velocity and angular acceleration are $ \omega $ and $ \alpha $ respectively. The linear speed, centripetal acceleration, and tangential acceleration of the point can be expressed as:

	Linear speed	Centripetal acceleration	Tangential acceleration
$ (a) $	$ R\omega $	$ R\omega^{2} $	$ R\alpha $
$ (b) $	$ R\omega $	$ R\alpha $	$ R\omega^{2} $
$ (c) $	$ R\omega^{2} $	$ R\alpha $	$ R\omega $
$ (d) $	$ R\omega $	$ R\omega^{2} $	$ R\omega $
$ (e) $	$ R\omega^{2} $	$ R\alpha $	$ R\omega^{2} $

Centripetal Acceleration, Rotational Kinematics, Rotational Motion

Intermediate

Mathematical

MCQ

The system above is NOT balanced since $m_2$ is twice the mass of $m_1$. Which of the following changes would NOT balance the system so that there is 0 net torque? Assume the plank has no mass of its own.

Rotational Motion, Static Torque, Torque

Intermediate

Conceptual

MCQ

A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously from the top of an inclined plane and do not slip, which one will reach the bottom first?

Rolling, Rotational Energy, Rotational Inertia, Rotational Motion

Advanced

Conceptual

A man with mass $ m $ is standing on a rotating platform in a science museum. The platform can be approximated as a uniform disk of radius $ R $ that rotates without friction at a constant angular velocity $ \omega $. Two students are discussing what the man should do if he wishes to change the angular velocity of the platform.

Student A says that the man should run towards the center of the platform, because this will decrease the moment of inertia of the man-platform system. Since $ L \propto I $, the angular momentum will decrease proportionately and the platform will slow down.

Student B says that since the platform is rotating counterclockwise, the man should run in a clockwise direction to slow the platform down. His feet will exert a frictional torque on the platform, which will cause an angular acceleration of the man-platform system.

Explain what is correct and incorrect about each students statement if anything.

Angular Momentum, Rotational Motion, Torque

Intermediate

Mathematical

The driver of a car traveling at $ 30.0 $ $ \text{m/s} $ applies the brakes and undergoes a constant negative acceleration of $ 2.00 $ $ \text{m/s}^2 $. How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of $ 0.300 $ $ \text{m} $?

Rolling, Rotational Kinematics

Intermediate

Mathematical

A 0.72-m-diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 N·m which accelerates it uniformly from rest through a total of 160 revolutions in 15.0 s. What is the mass of the sphere?

Rotational Kinematics, Torque

Advanced

Mathematical

FRQ

A pulley system consists of two blocks of mass $ 5 $ $ \text{kg} $ and $ 10 $ $ \text{kg} $, connected by a rope of negligible mass that passes over a pulley of radius $ 0.1 $ $ \text{m} $ and mass $ 2 $ $ \text{kg} $. The pulley is free to rotate about its axis. The system is released from rest, and the block of mass $ 10 $ $ \text{kg} $ starts to move downwards. Assume the pulley has a frictional force of $5.7$ Newtons acting on the outer edge of the pulley.

Atwood Machine, Dynamic Torque, Multi-Body Systems, Rotational Motion, Tension

Beginner

Mathematical

MCQ

A force of $17 \, \text{N}$ is applied to the end of a $0.63 \, \text{m}$ long torque wrench at an angle $45^\circ$ from a line joining the pivot point to the handle. What is the magnitude of the torque about the pivot point produced by this force?

Torque

Beginner

Mathematical

MCQ

An object’s angular momentum changes by $ 10 \,\text{kg} \cdot \text{m}^2/\text{s} $ in $ 2.0 $ $ \text{s} $. What magnitude average torque acted on this object?

Angular Momentum, Rotational Inertia, Rotational Motion, Torque

Advanced

Mathematical

MCQ

A meter stick with a uniformly distributed mass of $0.5 \, \text{kg}$ is supported by a pivot placed at the $0.25 \, \text{m}$ mark from the left. At the left end, a small object of mass $1.0 \, \text{kg}$ is placed at the zero mark, and a second small object of mass $0.5 \, \text{kg}$ is placed at the $0.5 \, \text{m}$ mark. The meter stick is supported so that it remains horizontal, and then it is released from rest. Find the change in the angular momentum of the meter stick, one second after it is released.

Angular Momentum, Rotational Motion, Torque

$\frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1})$
$\frac{h}{R}$

	Linear speed	Centripetal acceleration	Tangential acceleration
\( (a) \)	\( R\omega \)	\( R\omega^{2} \)	\( R\alpha \)
\( (b) \)	\( R\omega \)	\( R\alpha \)	\( R\omega^{2} \)
\( (c) \)	\( R\omega^{2} \)	\( R\alpha \)	\( R\omega \)
\( (d) \)	\( R\omega \)	\( R\omega^{2} \)	\( R\omega \)
\( (e) \)	\( R\omega^{2} \)	\( R\alpha \)	\( R\omega^{2} \)

Kinematics	Forces
\(\Delta x = v_i t + \frac{1}{2} at^2\)	\(F = ma\)
\(v = v_i + at\)	\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)	\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)	\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\)

Circular Motion	Energy
\(F_c = \frac{mv^2}{r}\)	\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)	\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)	\(KE_i + PE_i = KE_f + PE_f\)
	\(W = Fd \cos\theta\)

Momentum	Torque and Rotations
\(p = mv\)	\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)	\(I = \sum mr^2\)
\(p_i = p_f\)	\(L = I \cdot \omega\)

Simple Harmonic Motion	Fluids
\(F = -kx\)	\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)	\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)	\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)	\(F_b = \rho V g\)
\(a = -\omega^2 x\)	\(A_1v_1 = A_2v_2\)

Step	Derivation/Formula	Reasoning
1	$$ T_{1} – m_{1}g = m_{1}a $$	This is Newton’s second law for mass \(m_{1}\) moving upward.
2	$$ m_{2}g – T_{2} = m_{2}a $$	This is Newton’s second law for mass \(m_{2}\) moving downward.
3	$$ T_{1} = m_{1}g + m_{1}a \quad \text{and} \quad T_{2} = m_{2}g – m_{2}a $$	Rearrange the equations to solve for the tensions in the string.
4	$$ (T_{2} – T_{1})R = I\left(\frac{a}{R}\right) $$	This relates the net torque on the pulley to its moment of inertia \(I\) using the no‐slip condition \(\alpha = \frac{a}{R}\).
5	$$ T_{2} – T_{1} = \frac{I\,a}{R^{2}} $$	Simplify the torque equation by dividing both sides by \(R\).
6	$$ (m_{2}g – m_{2}a) – (m_{1}g + m_{1}a) = $$$$(m_{2}-m_{1})g – (m_{2}+m_{1})a =$$$$\frac{I\,a}{R^{2}} $$	Substitute the expressions for \(T_{1}\) and \(T_{2}\) into the torque equation.
7	$$ I = \frac{R^{2}}{a}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})a\Bigr] $$	Rearrange the equation to solve for the moment of inertia \(I\).
8	$$ h = \frac{1}{2}at^{2} $$	Use the kinematics relation for the heavy mass \(m_{2}\) falling a distance \(h\) from rest.
9	$$ a = \frac{2h}{t^{2}} $$	Solve for the acceleration \(a\) from the kinematics equation.
10	$$ I = \frac{R^{2}}{\frac{2h}{t^{2}}}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})\frac{2h}{t^{2}}\Bigr] $$	Substitute \(a = \frac{2h}{t^{2}}\) into the expression for \(I\).
11	$$ I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1}) $$	Simplify the expression to obtain \(I\) solely in terms of \(m_{1}, m_{2}, R, h, t\) and \(g\).
12	$$ \boxed{I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1})} $$	This is the final algebraic expression for the pulley’s moment of inertia.

Constant	Description
[katex]g[/katex]	Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]	Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]	Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]	Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]	Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]	Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]	Mass of the Sun

Variable	SI Unit
[katex]s[/katex] (Displacement)	[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)	[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)	[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)	[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)	[katex]\text{kilograms (kg)}[/katex]

Variable	Derived SI Unit
[katex]F[/katex] (Force)	[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)	[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)	[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)	[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)	[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)	[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)	[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)	[katex]\text{hertz (Hz)}[/katex]

Prefix	Symbol	Power of Ten
Pico-	p	[katex]10^{-12}[/katex]
Nano-	n	[katex]10^{-9}[/katex]
Micro-	µ	[katex]10^{-6}[/katex]
Milli-	m	[katex]10^{-3}[/katex]
Centi-	c	[katex]10^{-2}[/katex]
Deci-	d	[katex]10^{-1}[/katex]
(Base unit)	–	[katex]10^{0}[/katex]
Deca- or Deka-	da	[katex]10^{1}[/katex]
Hecto-	h	[katex]10^{2}[/katex]
Kilo-	k	[katex]10^{3}[/katex]
Mega-	M	[katex]10^{6}[/katex]
Giga-	G	[katex]10^{9}[/katex]
Tera-	T	[katex]10^{12}[/katex]

Self Paced. 100+ hours of lessons, worked examples, and targeted practice.

A custom-built platform to simplify and accelerate your learning journey.

Everything you need to master Physics — for free.

Meet Phy. The world's most advanced AI helper.

Collection of every AP FRQ with rapid, rubric aligned grading.

Find, try, and create 100s targeted quizzes. For Pro level students.

2000+ Physics problems & explanations to supercharge learning.

8 AP Physics units condensed into 8 short review guides.

Proven tips, tricks, and hacks to help you score that coveted 5.

Learn Physics, from scratch, at your own pace.

Guaranteed performance boost in scores and understanding.

5 weeks for a 5. Expedited progress. Stellar Results.

100+ hours of guided instruction to simplify physics & boost grades

Other 1-to-1 and group programs to accelerate Physics mastery.

New Tool FRQ Atlas - Find, Solve, and Grade Any FRQ In Seconds.

AP Physics

Unit 6 - Rotational Motion

FRQ

Mathematical

Advanced

Pro

Pro

Educator

Upgrade For More Credits

Part (a): Moment of Inertia

Part (b): Pulley Rotation

Need Help? Ask Phy To Explain

NEW UBQ QUIZ LAB

100s of AP aligned questions and quizzes to help you get a 5 even faster. Full Mock exams with Auto Grading and Adaptive explanations. Try out Nerd Notes', state of the art, quiz platform.

Topics in this question

Go Pro to remove ads + unlimited access to our AI learning tools.

Related Questions

5 More similar questions to the one you just solved

Nerd Notes

Discover the world's best Physics resources

Continue with

Error Report

Sign In to View Your Questions

Share This Question

PRO TIER

$20

Physics Shouldn't Be Hard

📚 Predict Your AP Physics Exam Score

How to UBQ

1. UBQ Credits

2. Auto Track Stats

3. Point System

Phy helps customize your learning experience.

4. Question Explantions

Phy gives feedback.

4. Feedback

We use cookies to improve your experience. By continuing to browse on Nerd Notes, you accept the use of cookies as outlined in our privacy policy.