Question

A rigid body is rotating about a fixed axis is initially at rest at t=0 torque acting on a body produces an angular acceleration (Anticlock wise taken as +ive)

• A. for O<t<2s, it rotates clockwise and for t> 2s it will be rotating in anticlockwise
• B. at t=2s, the disc will be at rest
• C. for t> 0, the disc will may undergo angular retardation or angular acceleration.
• D. all of these

Answer 1

Answer: (D)

all of these

Answer 2

The angular acceleration of the rigid body is given by $\alpha(t) = At - B$, where A and B are positive constants. The body is initially at rest at $t=0$, so its initial angular velocity $\omega(0) = 0$.

The angular velocity $\omega(t)$ is the integral of the angular acceleration with respect to time: $\omega(t) = \int \alpha(t) dt = \int (At - B) dt = \frac{1}{2}At^2 - Bt + C$. Using the initial condition $\omega(0) = 0$: $\omega(0) = \frac{1}{2}A(0)^2 - B(0) + C = 0 \implies C = 0$. So, the angular velocity is $\omega(t) = \frac{1}{2}At^2 - Bt = t(\frac{1}{2}At - B)$.

The direction of rotation is determined by the sign of $\omega(t)$. Anticlockwise is taken as positive. $\omega(t) = 0$ when $t=0$ or $\frac{1}{2}At - B = 0 \implies t = \frac{2B}{A}$. Let $t_0 = \frac{2B}{A}$. Since A and B are positive, $t_0 > 0$. For $0 < t < t_0$, $\frac{1}{2}At - B < 0$, so $\omega(t) < 0$. This means the rotation is clockwise. For $t > t_0$, $\frac{1}{2}At - B > 0$, so $\omega(t) > 0$. This means the rotation is anticlockwise.

Let's analyze the given options:

1. "for O<t<2s, it rotates clockwise and for t> 2s it will be rotating in anticlockwise" This statement describes the motion if the time at which the rotation changes direction is $t=2s$. From our analysis, the direction changes at $t = t_0 = \frac{2B}{A}$. So, this option is true if and only if $\frac{2B}{A} = 2$, which implies $B=A$.

2. "at t=2s, the disc will be at rest" The disc is at rest when $\omega(t) = 0$. We found that $\omega(t) = 0$ at $t=0$ and $t = \frac{2B}{A}$. So, the disc is at rest at $t=2s$ if and only if $\frac{2B}{A} = 2$, which implies $B=A$.

3. "for t> 0, the disc will may undergo angular retardation or angular acceleration." Angular acceleration is $\alpha(t) = At - B$. Angular retardation occurs when $\alpha$ and $\omega$ have opposite signs. Angular acceleration occurs when $\alpha$ and $\omega$ have the same sign. $\alpha(t) = 0$ when $t = \frac{B}{A}$. Let $t_1 = \frac{B}{A}$. Since A and B are positive, $t_1 > 0$. For $0 < t < t_1$, $\alpha(t) < 0$ (clockwise acceleration). For $t > t_1$, $\alpha(t) > 0$ (anticlockwise acceleration).

   We know $\omega(t) < 0$ for $0 < t < t_0 = \frac{2B}{A}$ (clockwise rotation) and $\omega(t) > 0$ for $t > t_0$ (anticlockwise rotation). Note that $t_0 = 2t_1$.

   Consider the interval $0 < t < t_1$: $\alpha < 0$ and $\omega < 0$. Both are negative (clockwise). This means the angular speed is increasing in the clockwise direction. This is angular acceleration. Consider the interval $t_1 < t < t_0$: $\alpha > 0$ and $\omega < 0$. They have opposite signs. This means the magnitude of the angular velocity is decreasing. This is angular retardation. Consider the interval $t > t_0$: $\alpha > 0$ and $\omega > 0$. Both are positive (anticlockwise). This means the angular speed is increasing in the anticlockwise direction. This is angular acceleration.

   Since A and B are positive, $t_1 = B/A > 0$ and $t_0 = 2B/A > 0$. The intervals $(t_1, t_0)$ and $(t_0, \infty)$ exist for $t>0$. For $t>0$, the disc undergoes angular acceleration in $(0, t_1)$ and $(t_0, \infty)$, and angular retardation in $(t_1, t_0)$. Therefore, for $t>0$, the disc will undergo both angular retardation and angular acceleration at different times. The statement "may undergo" is true because it does undergo both. This statement is true for any positive values of A and B.

4. "all of these" This option implies that options 1, 2, and 3 are simultaneously true. Options 1 and 2 are true if and only if $A=B$. Option 3 is true for any positive A and B. If we consider the specific case where $A=B$, then $t_1 = B/A = 1$ and $t_0 = 2B/A = 2$. In this case, $\alpha(t) = A(t-1)$ and $\omega(t) = \frac{1}{2}At^2 - At = At(\frac{1}{2}t - 1)$.

   • For $0 < t < 2$, $\omega(t) < 0$ (clockwise). For $t > 2$, $\omega(t) > 0$ (anticlockwise). Option 1 is true.
   • At $t=2$, $\omega(2) = A(2)(\frac{1}{2}(2) - 1) = 2A(1-1) = 0$. The disc is at rest. Option 2 is true.
   • For $t>0$:
     • $0 < t < 1$: $\alpha < 0$, $\omega < 0$. Acceleration.
     • $1 < t < 2$: $\alpha > 0$, $\omega < 0$. Retardation.
     • $t > 2$: $\alpha > 0$, $\omega > 0$. Acceleration. The disc undergoes both acceleration and retardation for $t>0$. Option 3 is true.

Since there exists a case (when $A=B$) where options 1, 2, and 3 are all true, the option "all of these" is the correct answer. The question asks which statement(s) is/are true, and "all of these" is presented as a single choice. This structure usually means that if the conditions for "all of these" are met for some valid parameters, then that is the intended answer.