Grade 11 : Physics - Rotational Motion (Set 2)

Explanation:
Angular speed is defined as $\omega = \frac{\theta}{t}$.
Substitute $\theta = 10\pi\text{ rad}$ and $t = 5\text{ s}$:
$$\omega = \frac{10\pi}{5} = 2\pi\text{ rad/s}$$

Explanation:
Angular acceleration is given by $\alpha = \frac{\omega_f - \omega_i}{t}$.
Given $\omega_i = 0\text{ rad/s}$, $\omega_f = 20\text{ rad/s}$, and $t = 4\text{ s}$:
$$\alpha = \frac{20 - 0}{4} = 5\text{ rad/s}^2$$

Explanation:
Using Newton's second law for rotation, $\tau = I\alpha$.
Given $I = 2kg\cdotm^2$ and $\alpha = 3\text{ rad/s}^2$:
$$\tau = 2 \times 3 = 6\text{ N}\cdot\text{m}$$

Explanation:
The formula for moment of inertia of a uniform rod about its center axis is:
$$I = \frac{1}{12} M L^2$$
Substitute $M = 3kg$ and $L = 2m$:
$$I = \frac{1}{12} \times 3 \times 2^2 = \frac{12}{12} = 1\text{ kg}\cdot\text{m}^2$$

Explanation:
For a solid uniform disk, $I = \frac{1}{2} M R^2$.
Substitute $M = 4kg$ and $R = 0.5m$:
$$I = \frac{1}{2} \times 4 \times (0.5)^2 = 2 \times 0.25 = 0.5\text{ kg}\cdot\text{m}^2$$

Explanation:
Angular momentum is given by $L = I \omega$.
Substitute $I = 4kg\cdotm^2$ and $\omega = 5\text{ rad/s}$:
$$L = 4 \times 5 = 20\text{ kg}\cdot\text{m}^2/\text{s}$$

Explanation:
Rotational kinetic energy formula is $K_{rot} = \frac{1}{2} I \omega^2$.
Substitute $I = 2kg\cdotm^2$ and $\omega = 4\text{ rad/s}$:
$$K_{rot} = \frac{1}{2} \times 2 \times 4^2 = 16\text{ J}$$

Explanation:
Tangential velocity is related to angular velocity by $v = r \omega$.
Substitute $r = 0.4m$ and $\omega = 10\text{ rad/s}$:
$$v = 0.4 \times 10 = 4\text{ m/s}$$

Explanation:
Tangential acceleration is calculated as $a_t = r \alpha$.
Given $r = 0.5m$ and $\alpha = 4\text{ rad/s}^2$:
$$a_t = 0.5 \times 4 = 2\text{ m/s}^2$$

Explanation:
Radial acceleration is given by $a_c = \omega^2 r$.
Substitute $\omega = 3\text{ rad/s}$ and $r = 2m$:
$$a_c = 3^2 \times 2 = 9 \times 2 = 18\text{ m/s}^2$$

Explanation:
Torque for perpendicular force is $\tau = r F \sin(90^\circ) = r F$.
Substitute $r = 0.5m$ and $F = 20\text{ N}$:
$$\tau = 0.5 \times 20 = 10\text{ N}\cdot\text{m}$$

Explanation:
For a thin circular hoop or ring about its central axis, $I = M R^2$.
Given $M = 2kg$ and $R = 1.5m$:
$$I = 2 \times (1.5)^2 = 2 \times 2.25 = 4.5\text{ kg}\cdot\text{m}^2$$

Explanation:
The moment of inertia of a solid sphere about a diameter is $I = \frac{2}{5} M R^2$.
Substitute $M = 5kg$ and $R = 1m$:
$$I = \frac{2}{5} \times 5 \times 1^2 = 2\text{ kg}\cdot\text{m}^2$$

Explanation:
Radius of gyration is related to moment of inertia by $I = M k^2$, so $k = \sqrt{\frac{I}{M}}$.
Substitute $I = 8kg\cdotm^2$ and $M = 2kg$:
$$k = \sqrt{\frac{8}{2}} = \sqrt{4} = 2\text{ m}$$

Explanation:
By conservation of angular momentum, $I_1 \omega_1 = I_2 \omega_2$.
Given $I_1 = 6kg\cdotm^2$, $\omega_1 = 2\text{ rad/s}$, and $I_2 = 3kg\cdotm^2$:
$$\omega_2 = \frac{6 \times 2}{3} = 4\text{ rad/s}$$

Explanation:
Work done by torque is given by $W = \tau \theta$.
Substitute $\tau = 5\text{ N}\cdotm$ and $\theta = 10\text{ rad}$:
$$W = 5 \times 10 = 50\text{ J}$$

Explanation:
Rotational power is given by $P = \tau \omega$.
Substitute $\tau = 8\text{ N}\cdotm$ and $\omega = 5\text{ rad/s}$:
$$P = 8 \times 5 = 40\text{ W}$$

Explanation:
Using rotational kinematic equation $\omega_f^2 = \omega_i^2 + 2\alpha\theta$:
With $\omega_i = 0$, $\alpha = 2\text{ rad/s}^2$, and $\theta = 9\text{ rad}$:
$$\omega_f^2 = 0 + 2(2)(9) = 36 \implies \omega_f = 6\text{ rad/s}$$

Explanation:
For a solid cylinder rolling without slipping, total kinetic energy is:
$$K_{total} = K_{trans} + K_{rot} = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{3}{4} M v^2$$
Substitute $M = 2kg$ and $v = 2m/s$:
$$K_{total} = \frac{3}{4} \times 2 \times 2^2 = \frac{3}{4} \times 8 = 6\text{ J}$$

Explanation:
The center of mass for two particles is $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.
Substitute given values:
$$x_{cm} = \frac{(1 \times 0) + (3 \times 4)}{1 + 3} = \frac{12}{4} = 3\text{ m}$$