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

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Mathematical

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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.

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Dynamic Torque, Rotational Motion, Torque

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

System	Rotational Inertia	Angular Speed
A	\( I_0 \)	\( \omega_0 \)
B	\( I_0 \)	\( 4\, \omega_0 \)
C	\( 2 I_0 \)	\( 2\, \omega_0 \)
D	\( 6 I_0 \)	\( \omega_0 \)

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]

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AP Physics

Unit 6 - Rotational Motion

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Part (a): Moment of Inertia

Part (b): Pulley Rotation

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