University

Physics 1

Physics 1

Fall 2024

Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A.

- What is the relationship between the coins’ magnitude of velocity, \(v_A\) and \(v_B\)?
- What is the relationship between the coins’ angular velocity, \(\omega_A\) and \(\omega_B\)?

- \(2v_A = v_B\)
- \(\omega_A = \omega_B\)

Rank, in order from least to greatest, the rotational inertia about the center of the rod for each of the following objects. The connecting rod has negligible mass.

\(I_a = m(\frac{L}{2})^2 + m(\frac{L}{2})^2 = \frac{1}{2}mL^2\)

\(I_b = 2m(\frac{L}{4})^2 + m(\frac{L}{4})^2 = \frac{3}{16}mL^2\)

\(I_c = \frac{m}{2}L^2 + \frac{m}{2}L^2 = mL^2\)

\(I_b \lt I_a \lt I_c\)

A uniform rectangular object has dimensions \(a > c \gg b\) . Consider the moment of inertia of the object about an axis through its center of mass, parallel to the \(x\), \(y\) and \(z\) axes.

Rank, from smallest to largest, the inertias \(I_x\), \(I_y\) and \(I_z\).

Inertia is smallest when the mass is closest to the axis.

\(I_x \lt I_z \lt I_y\)

Rank, in order from least to greatest, the kinetic energies of the following:

- a particle of mass \(M\) moving in a circle of radius \(R\) with angular velocity \(\omega\)
- a solid sphere of mass \(M\) and radius \(R\) spins with angular velocity \(\omega\)
- a hollow sphere of mass \(M\) and radius \(R\) spins with angular velocity \(\omega\)
- a solid disk of mass \(M\) and radius \(R\) spins with angular velocity \(\omega\) about its axis of symmetry

You may refer to the table of rotational inertias.

\(K_b \lt K_d \lt K_c \lt K_a\)

\(K= \frac{1}{2} I \omega^2\).

Since they all have the same \(\omega\), we just need to rank the rotational inertias. For the particle, \(I_a= MR^2\). The rest may be found on the table:

\(I_b= \frac{2}{5} MR^2\)

\(I_c= \frac{2}{3} MR^2\)

\(I_d= \frac{1}{2} MR^2\)

A wrench floats weightlessly in space. It is subjected to the same pair of equal and opposite forces, but at different locations on the wrench.

a

b

c

Rank the translational acceleration (\(a\)) and rotational acceleration (\(\alpha\)) in each case.

In all cases \(\vec{F}_{\rm{net}} = 0\), so \(a = 0\).

For rotation, \(\tau = Fd\), where \(d\) is the perpendicular distance between the force lines of action. And \(\tau = I \alpha\).

So: \(\alpha_a > \alpha_b > \alpha_c\)

a

b

c

A ball is rolling back and forth inside a bowl, between points \(A\) and \(E\). Give the sign (\(+\), \(-\), or \(0\)) of \(\omega\) and \(\alpha\) at each of the labeled points. Assume the ball is moving to the right at points \(B\), \(C\) and \(D\).

(Consider the rotation from rolling only, not the circular motion due to the arc of the bowl.)

A. \(\omega\) is \(0\), \(\alpha\) is \(-\)

B. \(\omega\) is \(-\), \(\alpha\) is \(-\)

C. \(\omega\) is \(-\), \(\alpha\) is \(0\)

D. \(\omega\) is \(-\), \(\alpha\) is \(+\)

E. \(\omega\) is \(0\), \(\alpha\) is \(+\)

To get the sign of \(\alpha\), consider the sign of \(\omega\) a moment before the point of interest and a moment after, then determine the sign of \(Delta \omega\).

\(\alpha\) will always have the same sign as \(\Delta \omega\) (as long as the time interval is small). For example at point \(A\), \(\omega\) was postive while rolling up and negative rolling down, so \(\Delta \omega = \omega_{final} - \omega_{initial} = (-) - (+)\), which is negative.

The following figures show a rotating wheel. Determine the directions of the vectors \(\vec \omega\) and \(\vec \alpha\).

A sphere of uniform density (with mass \(m\) and radius \(r\)) has a moment of inertia \(I = C m r^2\), where \(C=2/5\). The quantity \(C\) is known as the “moment of inertia factor”.

The Earth differs from a uniform sphere in two principle ways:

- The Earth’s core is much more dense than the rest of its mass.
- The Earth’s radius at the equator is greater than its radius at the poles.

Consider each of these differences individually. Do they tend to increase or decrease the value of \(C\) for the Earth?

A smaller value of \(C\) means less inertia: the object is easier to spin up.

- decrease – it’s easier to spin up a sphere when most of its mass is near the center
- increase – it’s harder to spin up a disk than a sphere

In fact, for the Earth, \(C=0.3307 \lt 2/5\). The dense core has a greater effect on \(I\) than the equatorial bulge.

A phonograph turntable is rotating at 33⅓ rpm (33⅓ rev/min), when a brake is applied to slow and stop it. The turntable has a rotational inertia of 0.2 kg·m². The magnitude of the decelerating torque applied by the brake increases with time, as \(\tau = 0.3 t\), with all quantities in MKS units.

Determine the time it takes for the turntable to come to rest.

We can find the angular acceleration from the torque: \(\tau = I \alpha = -At\), where the negative is used for angular deceleration, and \(A=0.3\) N·m/s.

\(\displaystyle \alpha(t) = -\frac{At}{I}\)

\(\displaystyle \Delta \omega = 0- \omega_0 =\int_0^{t_f} \alpha \; dt = -\frac{A}{I} \int_0^{t_f} t \; dt = -\frac{1}{2} \frac{A}{I} t_f^2\)

Solving for \(t_f\):

\(\displaystyle t_f = \sqrt{ \frac{2I \omega_0}{A}}\)

For greater confidence in this result, it’s worth taking a moment to check dimensionality.

With

\([I] = M L^2\),

\([A] = [F]L/T = MLT^{-2} \cdot L/T = ML^2/T^3\), and

\([\omega] = T^{-1}\), we have\(\displaystyle \left [ \sqrt{ \frac{2I \omega_0}{A}} \right ] = \sqrt{ ML^2 \cdot \frac{1}{T} \cdot \frac{T^3} {ML^2}} = \sqrt{T^2 } = T\)

This is a time, as we hoped.

Before we plug in values, get the initial angular velocity in MKS units:

\[ \omega_0 = 33⅓ \frac{\rm{rev}}{\rm{min}} \cdot \frac{2\pi \; \rm{rad}}{\rm{rev}} \cdot \frac{ 1 \; \rm{min}}{ 60 \; \rm{s}} = 3.491 \frac{\rm{rad}}{\rm{s}} \]

*Answer:*

\(\displaystyle t_f = \sqrt{ \frac{2(0.2)(3.491)}{0.3}} = 2.157\) s.

A uniform rod of mass \(m\) and length \(L\) is held vertically by two strings of negligible mass, as shown below.

- Immediately after the string is cut, what is the linear acceleration of the free end of the stick?
- At this time, what is the linear acceleration of the middle of the stick?

Express your answers in terms of \(g\) and the variables given.

You may refer to the table of rotational inertias.

Pivot is at the left end of rod. So \(I = \frac{1}{3} mL^2\).

Torque due to gravity at center is \(\tau = mg \frac{L}{2} = I \alpha\)

\(\displaystyle \alpha = mg \frac{L}{2I} = mg \frac{3L}{2mL^2} = \frac{3g}{2L}\)

- acceleration of stick end: \(a = L \alpha = \frac{3}{2} g\)
- acceleration of middle: \(a = \frac{L}{2} \alpha = \frac{3}{4} g\)

A pulley of moment of inertia 2.0 kg·m² is mounted on a wall as shown in the figure. Light strings are wrapped around two circumferences of the pulley and weights are attached.

Assume the following data: \(r_1=50\) cm, \(r_2=20\) cm, \(m_1=1.0\) kg, \(m_2=2.0\) kg.

- Find the angular acceleration of the pulley.
- Determine the linear acceleration of the weights.

Net torque is \(\tau = m_1 g \; r_1 - m_2 g \; r_2 = I \alpha\)

- \(\displaystyle \alpha = \frac{0.98}{2.0} = 0.49\) rad/s²
- \(a_1 = \alpha r_1 = 0.245\) m/s²
and

\(a_2 = \alpha r_2 = 0.098\) m/s²

*Last modified: Thu October 24 2024, 01:57 PM.*