Problem 16: A solid sphere of mass 5 kg is rolled upwards on an inclined plane with an angle of 30 degrees, starting with an initial velocity of 10 m/s. What is the maximum distance it travels up the inclined plane? Problem 17: A torque of 200 Nm is applied tangentially to the surface of a flywheel, which is initially at rest. It achieves an angular displacement of 100 rad in 5 seconds. What is the moment of inertia of the flywheel?

1. Problem Description

Problem 16: A solid sphere of mass 5 kg is rolled upwards on an inclined plane with an angle of 30 degrees, starting with an initial velocity of 10 m/s. What is the maximum distance it travels up the inclined plane?

Problem 17: A torque of 200 Nm is applied tangentially to the surface of a flywheel, which is initially at rest. It achieves an angular displacement of 100 rad in 5 seconds. What is the moment of inertia of the flywheel?

2. Solution Steps

Problem 16:

First, let's identify the given values:

$m = 5 \text{ kg}$ (mass of the sphere)

$\theta = 30^{\circ}$ (angle of the inclined plane)

$v_0 = 10 \text{ m/s}$ (initial velocity)

$v_f = 0 \text{ m/s}$ (final velocity at the maximum height)

The acceleration due to gravity acting along the inclined plane is $g \sin\theta$. Since the sphere is rolling up the incline, the linear acceleration $a$ is related to the angular acceleration $\alpha$ by $a = r\alpha$. For rolling without slipping, the torque is related to the moment of inertia $I$ and angular acceleration by $\tau = I\alpha$. For a solid sphere, the moment of inertia is $I = \frac{2}{5}mr^2$.

The net force along the inclined plane is $mg \sin \theta + F_f = ma$, where $F_f$ is the friction force. The torque about the center is $F_f r = I \alpha = \frac{2}{5} mr^2 \alpha$. Since $a = r\alpha$, we have $F_f r = \frac{2}{5} mr a$, so $F_f = \frac{2}{5} ma$.

Substituting into the force equation: $mg \sin \theta + \frac{2}{5} ma = ma$. Then $mg \sin \theta = \frac{3}{5} ma$, so $a = \frac{5}{3} g \sin \theta = \frac{5}{3} \times 9.8 \times \sin 30^{\circ} = \frac{5}{3} \times 9.8 \times 0.5 = \frac{4.9 \times 5}{3} = \frac{24.5}{3} \approx 8.17 \text{ m/s}^2$.

We can use the kinematic equation $v_f^2 = v_0^2 + 2as$ to find the distance $s$, where $v_f = 0$. Then $0 = v_0^2 + 2(-a)s$, which means $s = \frac{v_0^2}{2a} = \frac{10^2}{2 \times 8.17} = \frac{100}{16.34} \approx 6.12 \text{ m}$.

Since $7$ m is closest, we choose $7$m.

Problem 17:

Given:

$\tau = 200 \text{ Nm}$ (torque)

$t = 5 \text{ s}$ (time)

$\theta = 100 \text{ rad}$ (angular displacement)

Initial angular velocity $\omega_0 = 0$.

We can use the equation $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$ to find the angular acceleration $\alpha$.

$100 = 0 \times 5 + \frac{1}{2} \alpha (5^2) = \frac{25}{2} \alpha$.

Therefore, $\alpha = \frac{100 \times 2}{25} = 8 \text{ rad/s}^2$.

The torque is related to the moment of inertia $I$ and angular acceleration $\alpha$ by the equation $\tau = I \alpha$.

Therefore, $I = \frac{\tau}{\alpha} = \frac{200}{8} = 25 \text{ kgm}^2$.

3. Final Answer

Problem 16: v. 7 m

Problem 17: i. 25 kgm²