Grade 11 : Physics - Rotational Motion (Set 1)

Explanation: Angular velocity is defined as angular displacement per unit time $\omega = \frac{d\theta}{dt}$. The SI unit of angular displacement is radians (rad) and time is seconds ($s$), so the unit is $rad/s$.

Explanation: Linear distance arc length $s = r\theta$. Differentiating both sides with respect to time gives $\frac{ds}{dt} = r \frac{d\theta}{dt}$, which leads directly to $v = r\omega$.

Explanation: Torque is defined as the cross product $\vec{\tau} = \vec{r} \times \vec{F}$. The magnitude of vector cross product is $|\vec{\tau}| = r F \sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$.

Explanation: For a thin ring, all mass elements are situated at an equal distance $R$ from the central axis. Integrating $I = \int r^2 dm = R^2 \int dm = M R^2$.

Explanation: Integrating concentric thin rings of a solid disc yields the standard formula $I = \frac{1}{2} M R^2$.

Explanation: Just as linear force $F = m a$, rotational torque $\tau = I \alpha$, where $I$ is moment of inertia and $\alpha$ is angular acceleration.

Explanation: Analogous to translational kinetic energy $K = \frac{1}{2} m v^2$, rotational kinetic energy is $K_{rot} = \frac{1}{2} I \omega^2$.

Explanation: Analogous to linear momentum $p = mv$, angular momentum for a rotating rigid body is $L = I \omega$.

Explanation: Convert revolutions per minute to radians per second: $\omega = 120 \times \frac{2\pi}{60} = 4\pi\text{ rad/s}$.

Explanation: Setting origin at 2 kg mass: $x_{cm} = \frac{m_1 (0) + m_2 (d)}{m_1 + m_2} = \frac{3 \times 10}{2 + 3} = \frac{30}{5} = 6m$.

Explanation: Moment of inertia $I = \text{mass} \times (\text{distance})^2$. Hence, dimensions are $[M^1 L^2 T^0]$.

Explanation: By the law of conservation of angular momentum, $\tau_{ext} = \frac{dL}{dt}$. If $\tau_{ext} = 0$, then $L = \text{constant}$.

Explanation: Moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5} M R^2 = \frac{2}{5} \times 5 \times (2)^2 = 8kg\cdotm^2$.

Explanation: Using kinematic formula $\omega = \omega_0 + \alpha t \Rightarrow 20 = 0 + \alpha (4) \Rightarrow \alpha = 5\text{ rad/s}^2$.

Explanation: Pulling arms inward decreases moment of inertia $I$. By conservation of angular momentum ($L = I\omega = \text{constant}$), a decrease in $I$ causes $\omega$ to increase.

Explanation: Analogous to linear work $W = F s$, rotational work done by constant torque is $W = \tau \theta$.

Explanation: Since force is perpendicular ($\theta = 90^\circ$), $\tau = r F = 0.5 \times 10 = 5.0\text{ N}\cdotm$.

Explanation: Moment of inertia $I = \frac{1}{12} M L^2 = M k^2 \Rightarrow k = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$. Thus, both B and C represent the same value.

Explanation: By definition of rotational kinematics in vector notation, $\vec{v} = \vec{\omega} \times \vec{r}$. Order matters because vector cross products are anti-commutative.

Explanation: Angular momentum is calculated as $L = I \omega = 4 \times 10 = 40kg\cdotm^2/\text{s}$.