Rotational Motion

Ans. (a) $\frac{m}{M} = \frac{1}{4}$ ; (b) $x = \frac{2L}{3}$ ; (c) $\frac{v_0}{2\sqrt{2}}$
(a) $e = \frac{v^{\prime}_{2} - v^{\prime}_{1}}{v_{1} - v_{2}} = \frac{\frac{\omega L}{2} + v_{cm}}{v_0}$
$v _ {C M} = \frac {m v _ {0}}{M} \mathrm{and} m v _ {0} \frac {L}{2} = \frac {m _ {1} L ^ {2}}{1 2} \omega$

$\omega = \frac {6 m v _ {0}}{M L}$
$\frac {\frac {3 m v _ {0}}{M} + \frac {m v _ {0}}{M}}{v _ {0}} = 1$
$\Rightarrow \frac {m}{M} = \frac {1}{4}$
(b) $v_{cm} - \omega x = 0$
$\therefore x = \frac {L}{6}$
$\therefore \frac {L}{2} + \frac {L}{6} = \frac {2 L}{3}$
(c) $\omega \ell = \frac{3}{2} v_0$

$\omega t = \frac {\omega \pi \ell}{3 v _ {0}}$
$\frac {\pi \ell}{3 v _ {0}} \times \frac {3 v _ {0}}{2 \ell} = \frac {\pi}{2}$
$\therefore \text {After} \frac {\pi}{2}$

Ans. 75 J
$J = 1 0 \times 1 = I \omega - 0$
$I = \frac {1}{3} m \ell^ {2}$
$\therefore \omega = 1 5$

$\therefore K E = \frac {1}{2} I \omega^ {2} = \frac {1}{2} \times 2 \times \frac {1 ^ {2}}{3} \times 1 5 ^ {2} = 7 5 J$

Ans. 48
Let $V$ be the linear velocity of centre of rod which is struck by a horizontal force. For translational motion, $P_0 = MV_{\mathrm{cm}}$
For rotational motion, $P_{0}x = I_{\mathrm{R}}\omega$
Angular velocity of rod $\omega = \frac{P_0x}{ML^2} = \frac{3P_0x}{ML^2}$
Initial speed of centre of mass of the rod $\frac{\omega L}{2}=\frac{3P_{0}x}{2ML}$
Impulse delivered by the pivot is $I = mV_{\mathrm{cm}} - m\omega \frac{L}{2}$
Impulse is zero when $V_{cm} = \omega \frac{L}{2}$ ; $V_{cm} = \frac{P_0}{M} = \frac{3P_0x}{2ML} \Rightarrow x = \frac{2L}{3}$

Ans. 8
$m v _ {0} \frac {\ell}{4} = I _ {c m} \omega$
$m v _ {0} \frac {\ell}{4} = \frac {5}{2 4} m \ell^ {2} \omega$
$\omega = \frac {6}{5} \frac {v _ {0}}{\ell}\tag{... (i}$
$m v _ {0} = 2 m v _ {\mathrm{c}}$
$v _ {c} = \frac {v _ {0}}{2}\tag{... (ii}$
$v _ {P} = v _ {C} - \omega x = 0$
$x = \frac {v _ {c}}{\omega} = \frac {5 \ell}{1 2}$
Distance of P from A is $x_{0}=x+\frac{\ell}{4}$
$x _ {0} = \frac {8 \ell}{1 8} = 8 0 c m$
$\frac {x _ {0}}{1 0} = 8$

Ans. (3/8) $F_{g}$
$N _ {1} + P + f r _ {2} = F _ {g}$
... (1)
$N _ {2} = f r _ {1} = \mu N _ {1}$
$f r _ {1} = \mu N _ {1} \text { and } f r _ {2} = \mu N _ {2}$
$\tau \mathrm{about} O = 0$
$- P r + f r _ {1} r + f r _ {2} r = 0$
$\therefore P = f r _ {1} + f r _ {2} = \mu (N _ {1} + N _ {2})\tag{...(2}$

$\mathrm{From(1)} N _ {1} + P + \mu N _ {2} = F _ {g}\tag{...(3}$
$\therefore N _ {1} (1 + \mu^ {2}) = F _ {g} - P$
$\therefore N _ {1} = \frac {F _ {g} - P}{(1 + \mu^ {2})}$
$\begin{array}{c} \mathrm{Puttingin} P = \mu (N _ {1} + \mu N _ {1}) \\ = \mu N _ {1} (1 + \mu) \end{array}$
$P = \mu \frac {(F _ {g} - P) (1 + \mu)}{1 + \mu^ {2}}$
$\therefore P = \frac {\frac {1}{2} \left(1 + \frac {1}{2}\right)}{1 + \frac {1}{4}} \left(F _ {g} - P\right)$
$\Rightarrow 5 P = 3 F _ {g} - 3 P$
$\therefore P = \frac {3 F _ {g}}{8}$

Ans. (a) $\sqrt{3} ml\omega^2$ , (b) $(F_{net})_x = -\frac{F}{4}$ , $(F_{net})_y = \sqrt{3} ml\omega^2$
(a) $3mw^{2}\left(\frac{\sqrt{3}}{2}\ell \frac{2}{3}\right) = \sqrt{3} mw^{2}\ell$
(b) $F_{y} = \sqrt{3} m\ell \omega^{2}$
But for $F_{x}$
$F \frac {\sqrt {3}}{2} \ell = 2 m \ell^ {2} \alpha$
And $F - F_{x} = \frac{F}{4m} 3m$

Ans. (a) $a_{c}=\frac{4F}{3m_{1}+8m_{2}}$ , $a_{p}=2a_{c}$
(b) Friction at the top of the cylinder $= 3m_{1}F(3m_{1} + 8m_{2})$ towards right; friction at the bottom $= m_{1}F / (3m_{1} + 8m_{2})$ towards right. $F - f_{1} = M_{2}(a + \alpha r) = m_{2}(2a)$ $f_{1} - f_{2} = m_{1}a$ $(f_{1} + f_{2})r = \frac{m_{1}r^{2}}{2}\alpha = \frac{m_{1}a}{2}$ $f_{1} = \frac{3}{4} m_{1}a$ $\therefore F = 2m_{2}a + \frac{3}{4} m_{1}a$ $\therefore a = \frac{4F}{8m_2 + 3m_1}$

Ans. 6N, $-0.6\hat{j} \pm 0.6\hat{k}$
(a) $18 - f = 2a$
$f \times 0. 1 = \frac {2 (. 1) ^ {2}}{2} \alpha$
$\therefore f = 6 N$

(b) $\vec{r} = (-0.1\hat{j} - 0.1\hat{k})$ ; $f = 6\hat{i}$
$\therefore - 0. 6 (\hat {j} \times \hat {\imath}) - 0. 6 (\hat {k} \times \hat {j}) = 0. 6 \hat {k} - 0. 6 \hat {j}$

Ans. (a) 0.1 m, (b) 1 rad/s, (c) laminar plate will never come to rest.
For Laminar sheet $I_{AB} = I_{CM} + my^{2} = 1.2 + 30y^{2}$ Also impulse = $\Delta P$ $\therefore \quad 6 = m(V_{f} - V_{i}) \Rightarrow 6 = 30(w_{fy} - w_{i}y)$ $\Rightarrow \quad 6 = 30y(w_{f} - w_{i}) \Rightarrow 1 = 5y(w_{f} - w_{i})$
Also angular impulse = change in angular momentum $\Rightarrow$ $6 \times 0.5 = I_{AB} \left( w_f - w_i \right)$ [About AB] $\Rightarrow$ $w_f = \frac{3}{I_{AB}} - 1$ [Given $w_i = 1 \, rad/s$ ] $\Rightarrow$ $w_f = \frac{3}{1.2 + 30yz} - 1$ ...(2)
From (1) and (2) $l = 5y \left[ \frac{3}{1.2 + 30yz} \right] \Rightarrow 30y^2 - 15y + 1.2 = 0$ $\therefore y = 0.1 \text{ or } 0.4$ $\therefore w_f = 1 \, rad/s \text{ for } y = 0.1$ $= -0.5 \, rad/s \text{ for } y = 0.4 \text{ (Not possible as laminar plate turns back)}$ $\therefore \text{ location of COM} = 0.1 \, m$ $w_f = 1 \, rad/s \Rightarrow Laminar plate will never stop.$ Ans. (a) 0.1 m (b) 1 rad/s (c) laminar plate will never come to rest.
EXERCISE - JEE (Main) PYQ

Ans. (C)
$\tau_ {\mathrm{ICR}} = F R = I _ {\mathrm{ICR}} \alpha = \frac {5}{7} m R ^ {2} \alpha$
$\Rightarrow a _ {n} = \frac {5}{7} \frac {F}{m} = \frac {5}{7} \frac {a t}{m}$
$F - f = m a \Rightarrow f = \frac {2 a t}{7}$
Slips $\Rightarrow 7\mu_{s}mg / 2a$

Once it slips
$F - \mu m g = m a _ {\mathrm{n}}$
$a t - \mu m g = m a _ {\mathrm{n}}$
$a _ {n} = \frac {a t}{m} - \mu g$

Ans. (C)
$F r = I \alpha$
$F = \frac {1}{2} M R \alpha = 3 N$

Ans. (D)
Wrt COM of hemisphere
$\mathrm{Torque} = I _ {\mathrm{com}}. \alpha$
$F \cdot \frac {R}{2} = I _ {c o m} \alpha$
$F \cdot \frac {R}{2} = \left[ \frac {2}{3} M R ^ {2} - m \left(\frac {R}{2}\right) ^ {2} \right] \alpha$
On solving

$\alpha = \frac {6 F}{5 M R}$

Ans. (A)
$\left(\vec {\tau} _ {n e t}\right) _ {C} = 0$
$\frac {F 3 \ell}{4} - m g \frac {\ell}{4} \sin \alpha = 0$
$F = \frac {m g}{3} \sin \alpha$
$\mu N = f = m g \cos \alpha$
$N + F = m g \sin \alpha$
$\mu (m g \sin \alpha - F) = m g \cos \alpha$
$\mu = \frac {m g \cos \alpha}{m g \sin \alpha - F}$

$\mu = \frac {m g \cos \alpha}{m g \sin \alpha - \frac {m g \sin \alpha}{3}} = \frac {1}{\frac {2 \tan \alpha}{3}}$
$\mu = \frac {9}{8}$

Ans. (D)
$\begin{array}{r l} & {A M = m \vec {r} \times \vec {v} = 2 (+ i + j) \times (2) (+ i - j + k)} \\ & {\qquad = - 8 \hat {k} - 4 j + 4 i} \end{array}$
$\therefore \quad \text { About } z \text { axis } = - 8$

Ans. (B)
$L_{i} = mrv\sin (90 - \theta)$ $= mdv\cos \theta$ $L_{f} = -mdv\cos \theta$ $\therefore \Delta L = 2mdv\cos \theta$

Ans. (C)
Ans. (c) $I_{1}\omega_{1}^{2} = I_{2}\omega_{2}^{2}$ $\therefore \quad \omega_{1}^{2} = 4\omega_{2}^{2}$ $\therefore \quad \omega_{1} = 2\omega_{2}$ $\frac{L_{2R}}{L_R} = \frac{I_2\omega_2}{I_1\omega_1} = \frac{4}{1} \times \frac{1}{2} = 2$

Ans. (D)
Angular momentum of particle remains constant as it moves along the line BC.

Ans. (B)
$\begin{array}{r l} & {\mathrm{Ans.} (\mathbf {B})} \\ & {\vec {L} = \vec {r} \times m \vec {u}} \\ & {\quad = m [ (v _ {0} \cos \theta t) \hat {\imath} + (v _ {0} \sin \theta t - \frac {1}{2} g t ^ {2}) \hat {j} ] \times [ v _ {0} \widehat {\cos \imath} + (v _ {0} \sin \theta - g t) \hat {j} ]} \\ & {\quad = m [ \frac {t ^ {2} \cos \theta}{2} v _ {0} ] (- \hat {k})} \end{array}$

Ans. (A)

$L _ {0} = m u \frac {1}{2} g t ^ {2} - m (g t) u t = \frac {1}{2} m u g t ^ {2} - m u g t ^ {2} = - \frac {1}{2} m u g t ^ {2} \hat {k}$

Ans. (B)
Since no external torque is acting on the system $\tau_{ext} = 0 \Rightarrow \frac{dL}{dt} = 0$ here $L$ is angular momentum. $L = I\omega = \mathrm{constant}$ .
So, total angular momentum is constant.
Since system is not having any energy exchange. Total energy is also conserved.
Because Beads are going down, distance from axis of rotation increasing, there I is increasing. Since $I_{\omega}$ is constant, and I us increasing $\omega$ has to decrease.
So, Total angular momentum and total energy is constant.

Ans. (B)
Angular momentum will be conserved about O

$\therefore I = m | d - v t ] ^ {2} + I _ {\mathrm{disk}}$

Ans. (C) Let $\omega$ is the angular speed of disk $A$ after person jumps from it. $\frac{1}{2} MR^2\omega + m(u + R\omega)R = \frac{1}{2} MR^2\omega_0 + m\omega_0R^2$ $m(u + R\omega)R = \frac{1}{2} MR^2\omega_0$

Ans. (A)
Moment of inertia decreases therefore angular velocity increases. KE also increases and therefore work done is positive.

Ans. (D) $L_{i} = L_{f}$ $\Rightarrow m\omega R^2 +I\omega = mvR + I\omega '$ $Q\omega ' = \frac{(mR^2 + I)\omega - mvR}{I}$

Ans. (B) Conservation of angular momentum $L_{i} = L_{f} \Rightarrow \omega_{0} \times \frac{mR^{2}}{2} = \omega' \left[ \frac{mR^{2}}{2} + \frac{m\ell^{2}}{3} \right] \Rightarrow \omega' = \frac{3\omega_{0}R^{2}}{3R^{2} + 2\ell^{2}}$