Detailed Answer

The cylinder has 2 types of energy, kinetic \(\frac{1}{2}mv^2\) and rotational \(\frac{1}{2}\theta\omega^2\). Since we have a cylinder \(\theta = \frac{1}{2}mr^2\) and \(\omega = \frac{v}{r}\), plugging these in gives:

\[ \frac{1}{2} \cdot \frac{1}{2}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{4}mv^2 \]

Adding this to the kinetic energy gives: \[ \frac{3}{4}mv^2 \]

So, the correct answer is B.

Detailed Answer

The rod is solid, meaning all parts of the rod rotate the same angle, in the same amount of time, hence the angular speed is the same everywhere on the rod.

Detailed Answer

The final angular speed can simply be calculated with: \[ 5\;\frac{rad}{s^2} \cdot 5\;s = 25\;\frac{rad}{s} \]

The angular displacement is: \[ \Delta \Theta = \frac{1}{2}(5)(5)^2 = 62.5\;\pi \;rad \]

Detailed Answer

The arm of the force is by definition the shortest (perpendicular) distance from the force to the axis.

Detailed Answer

The torques in both directions around the pivot have to be equal, thus: \[ 50 \cdot 15 = 5 \cdot 50 + 25 \cdot L \] Solving this gives \( L = 20 \).

Detailed Answer

The weight force acts at the center of the beam vertically down, whereas the tension acts at the end of the rod vertically up. We can take the torque about the hinge to get: \[ T\cdot 4 = 98\cdot 2 \] Solving for \(T\): \[ T = 49N \]

Detailed Answer

Using Newton’s second law for rotation: \[ \tau = I \alpha \implies \alpha = \frac{\tau}{I} = \frac{10}{5.0} = 2.0 \, \text{rad/s}^2 \]

Detailed Answer

The rotational kinetic energy is given by: \[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \] Substituting the given values: \[ KE_{\text{rot}} = \frac{1}{2} \cdot 0.4 \cdot 30^2 = 180 \, \text{J} \]

Detailed Answer

The moment of inertia of a solid sphere rotating about its diameter is: \[ I = \frac{2}{5} M R^2 = \frac{2}{5} (2.0)(0.3)^2 = 0.072 \, \text{kg·m}^2 \]

Angular momentum is: \[ L = I \omega = 0.072 \cdot 5.0 = 0.36 \, \text{kg·m}^2/\text{s} \]

Detailed Answer

For a rolling sphere, the rotational kinetic energy is:

\[ E_{\text{rot}} = \frac{2}{7} E_{\text{t}} = \frac{2}{7} \cdot 50 = 20 \, \text{J} \]

This can be derived:

\[ E_t = E_K + E_{\text{rot}} \] \[ = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] \[ = \frac{1}{2}mv^2 + \frac{1}{2} \left(\frac{2}{5}mR^2\right) \left(\frac{v}{r}\right)^2 \] \[ = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 \] \[ = \frac{7}{10}mv^2 \]

Thus, \[ \frac{E_{\text{rot}}}{E_t} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{2}{7} \]

Detailed Answer

Using the equation for angular displacement: \[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \] \[ = 0 + \frac{1}{2} \cdot (4.0) \cdot (5.0)^2 \] \[ = 0 + \frac{1}{2} \cdot 4.0 \cdot 25 \] \[ = \frac{1}{2} \cdot 100 = 50 \, \text{rad} \] Convert radians to revolutions: \[ \text{Revolutions} = \frac{\theta}{2\pi} = \frac{50}{2\pi} \approx 7.96 \approx 8 \, \text{revolutions.} \]

Detailed Answer

Using conservation of angular momentum:

\[ L_{\text{initial}} = L_{\text{final}} \]

The initial angular momentum is:

\[ L_{\text{initial}} = I_1 \omega_1 = (0.5)(10) = 5.0 \, \text{kg·m}^2/\text{s} \]

The final angular momentum is:

\[ L_{\text{final}} = (I_1 + I_2) \omega_{\text{final}} \]

Equating:

\[ 5.0 = (0.5 + 1.0) \omega_{\text{final}} \implies \omega_{\text{final}} = \frac{5.0}{1.5} = 3.3 \, \text{rad/s} \]

Detailed Answer

The initial rotational kinetic energy is:

\[ KE_{\text{initial}} = \frac{1}{2} I \omega^2 = \frac{1}{2}(2.0)(10)^2 = 100 \, \text{J} \]

The brake does \(100 \, \text{J}\) of work on the system. The work-energy theorem states:

\[ W = KE_{\text{final}} - KE_{\text{initial}} \]

Substituting \(W = -100 \, \text{J}\) and \(KE_{\text{initial}} = 100 \, \text{J}\):

\[ -100 = KE_{\text{final}} - 100 \implies KE_{\text{final}} = 0 \]

If the final kinetic energy is zero, the angular velocity must also be zero:

\[ \omega_{\text{final}} = 0 \, \text{rad/s} \]

Detailed Answer

The torque on the disk is given by:

\[ \tau(t) = F(t) R = (10t)(0.4) = 4t \, \text{N·m} \]

The moment of inertia of the disk is:

\[ I = \frac{1}{2} M R^2 = \frac{1}{2}(5.0)(0.4)^2 = 0.4 \, \text{kg·m}^2 \]

Substitute \( I \) into the angular acceleration formula:

\[ \alpha(t) = \frac{\tau(t)}{I} \]

\[ \alpha(t) = \frac{4t}{0.4} = 10t \, \text{rad/s}^2 \]

The angular velocity is obtained by integrating \( \alpha(t) \):

\[ \omega(t) = \int \alpha(t) \, dt = \int 10t \, dt = 5t^2 + C \]

Since the disk starts from rest, \( \omega(0) = 0 \), so \( C = 0 \):

\[ \omega(t) = 5t^2 \]

At \( t = 2.0 \, \text{s} \):

\[ \omega(2.0) = 5(2.0)^2 = 20 \, \text{rad/s} \]