Rotational Motion

Ans. (D)
$\begin{array}{l} \text {Force on block along slot} = m \omega^ {2} r = m a = m \left(\frac {v d v}{d r}\right) \\ \int_ {0} ^ {v} v d v = \int_ {R / 2} ^ {r} \omega^ {2} r d r \\ \frac {v ^ {2}}{2} = \frac {\omega^ {2}}{2} \left(r ^ {2} - \frac {R ^ {2}}{4}\right) \\ \Rightarrow \quad v = \omega \sqrt {r ^ {2} - \frac {R ^ {2}}{4}} = \frac {d r}{d t} \\ \Rightarrow \quad \int_ {R / 4} ^ {r} \frac {d r}{\sqrt {r ^ {2} - \frac {R ^ {2}}{4}}} = \int_ {0} ^ {t} \omega d t \\ \ell n \left(\frac {r + \sqrt {r ^ {2} - \frac {R ^ {2}}{4}}}{\frac {R}{2}}\right) - \ell n \left(\frac {R / 2 + \sqrt {\frac {R ^ {2}}{4} - \frac {R ^ {2}}{4}}}{\frac {R}{2}}\right) = \omega t \\ \Rightarrow \quad r + \sqrt {r ^ {2} - \frac {R ^ {2}}{4}} = \frac {R}{2} e ^ {\omega t} \\ \Rightarrow \quad r ^ {2} - \frac {R ^ {2}}{4} = \frac {R ^ {2}}{4} e ^ {2 \omega t} + r ^ {2} - 2 r \frac {R}{2} e ^ {\omega t} \\ \Rightarrow \quad r = \frac {\frac {R ^ {2}}{4} e ^ {2 \omega t} + \frac {R ^ {2}}{4}}{\mathrm{Re} ^ {\omega t}} = \frac {R}{4} (e ^ {\omega t} + e ^ {- \omega t}) \end{array}$

Ans. (C)
$\vec {N} _ {1} = m g \hat {k}$
$\begin{array}{r l} & {\vec {N} _ {2} = 2 m (V _ {r o t} ^ {\prime} \times \vec {\omega}) \hat {j} = 2 m \left[ \frac {\omega R}{4} (e ^ {\omega t} - e ^ {- \omega t}) \right] \omega \hat {j}} \\ & {\quad = \frac {1}{2} m \omega^ {2} R (e ^ {\omega t} - e ^ {- \omega t}) \hat {j}} \end{array}$

Ans. (ABC)
When the bar makes an angle $\theta$ ; the height of its COM (mid point) is $\frac{L}{2}\cos\theta$
∴ Displacement = $\frac{L}{2} - \frac{L}{2}\cos\theta = \frac{L}{2}(1 - \cos\theta)$
Since force on COM is only along the vertical direction, hence COM is falling vertically downward. Instantaneous torque about point of contact is
$\tau = M g \times \frac {L}{2} \sin \theta$
Now, $x = \frac{L}{2}\sin \theta$
$y = L \cos \theta$
$\frac {x ^ {2}}{(L / 2) ^ {2}} + \frac {y ^ {2}}{L ^ {2}} = 1$

L/2 sin θ
Path of $A$ is an ellipse.

Ans. (BC or C)
If we assume that magnitude of force remain constant then option (C) would be correct and if we assume that force vector remain constant then option (BC) would be correct.

Ans. (BC)
$V = \frac {k r ^ {2}}{2}$
$F = - k r (\text { towards centre }) \left[ F = - \frac {d V}{d r} \right]$
$\mathrm{At} \qquad r = R,$
$kR=\frac{mv^{2}}{R}$ [Centripetal force]
$v = \sqrt {\frac {k R ^ {2}}{m}} = \sqrt {\frac {k}{m}} R$

$L = m \sqrt {\frac {k}{m}} R ^ {2} = \sqrt {m k} R ^ {2}$

Ans. (A)

For $\theta_{max}$ , the football is about to roll, then $N_{2}=0$ and all the forces (Mg and $N_{1}$ ) must pass through contact point
$\therefore \cos (9 0 ^ {\circ} - \theta_ {\mathrm{max}}) = \frac {r}{R}$
$\theta_ {\mathrm{max}} = \frac {r}{R}$
For no slipping at the ground,
$V_{centre} = \omega R$ (R is radius of roller)
∴ Velocity of scale = $(V_{center} + \omega r)$ [r is radius of axle]
Given, $V_{center} \cdot t = 50 \, cm$ ∴ Distance moved by scale = $(V_{center} + \omega r)t$ $=\left(V_{center}+\frac{V_{center}r}{R}\right)t=\frac{3V_{center}}{2}\cdot t$ = 75 cm

Ans. (1)
$\vec {F} = 4 \hat {\imath} - 3 \hat {j}$
$\vec {r} _ {1} = 5 \hat {\imath} + 5 \sqrt {3} \hat {j} \mathrm{and} \vec {r} _ {2} = - 5 \hat {\imath} + 5 \sqrt {3} \hat {j}$
Torque about 'O'
$\vec {\tau} _ {o} = \vec {r} _ {1} \times \vec {F} = (- 1 5 - 2 0 \sqrt {3}) \hat {k} = (1 5 + 2 0 \sqrt {3}) (- \hat {k})$
Torque about 'Q'
$\vec {\tau} _ {Q} = \vec {r} _ {2} \times \vec {F} = (- 1 5 + 2 0 \sqrt {3}) \hat {k} = (1 5 - 2 0 \sqrt {3}) (- \hat {k})$

Ans. (25.60)
Initially

$f _ {1 _ {K}} = \mu_ {K} N _ {1}$
$f _ {1 _ {K}} = 0. 3 2 N _ {1}$
$f _ {K _ {1}} = f _ {L _ {2}}$
$N _ {1} \mathbf {x} _ {L} = \frac {4 N _ {1}}{5}$
Now $x_{R}$ = distance when right finger stops and left finger starts moving
$\left(\tau_ {\mathrm{n}} = 0\right) _ {\mathrm{aboutcentre}}$
$\Rightarrow N _ {1} \mathrm{x} _ {L} = N _ {2} (\mathrm{x} _ {R})$
$f _ {L _ {1}} = f _ {K _ {2}}$
$\Rightarrow \quad 0. 4 N _ {1} = 0. 3 2 N _ {2}$
$5 N _ {1} = 4 N _ {2}$
$\frac {4 N _ {2}}{5} (3 2) = N _ {2} x _ {R}$
$x _ {R} = \frac {1 2 8}{5} = 2 5. 6 c m$

Ans. (BD)
$F - f = m a _ {c}$
$f R = I _ {c} \alpha$
$a _ {c} - \alpha R = 0$
$F - I _ {c} \frac {\alpha}{R} = m a _ {c}$
$a _ {c} = \frac {F}{\frac {I _ {c}}{R ^ {2}} + m}$
$f = \frac {I _ {c} \alpha}{R} = \frac {I _ {c}}{R ^ {2}} a _ {c} = \frac {I _ {c}}{R ^ {2}} \frac {F}{\left[ \frac {I _ {c}}{R ^ {2}} + m \right]}$

$f = \frac {F}{\left(m + \frac {I _ {C}}{R ^ {2}}\right)}$
Thin walled hollow cylinder,
$I _ {c} = m R ^ {2}$
$a _ {c} = \frac {F}{2 m}$
$f R = I _ {c} \alpha = \frac {I _ {c} a _ {c}}{R}$
$f = \frac {I _ {c} a _ {c}}{R ^ {2}} \leq \mu m g$
$a _ {c} \leq \frac {\mu m g R ^ {2}}{I _ {c}}$
For solid cylinder $I_{c}=\frac{mR^{2}}{2}$
$a _ {c} \leq 2 \mu g$
$\left(a _ {c}\right) _ {\mathrm{max}} = 2 \mu g$

Ans. (ABC)
$\vec {r} _ {A} = - \hat {j}$
$S = \frac {1}{2} a t ^ {2} = 2 0 m$
$\vec {\tau} _ {0} = \vec {r} \times \vec {F}; \vec {r} _ {B} = 1 0 \hat {\imath} - \hat {j}$
$\vec {F} = m \vec {a} = 0. 2 \times 1 0 \hat {\imath} = 2 \hat {\imath}$
$\vec {\tau} _ {0} = (1 0 \hat {\imath} - \hat {j}) \times (2 \hat {\imath})$
$\vec {\tau} _ {0} = 2 \hat {k}$
$\vec {L} _ {0} = \vec {r} _ {B} \times \vec {p} = \vec {r} _ {B} \times m \vec {v}$
$\vec {v} = \vec {a} t = 1 0 \hat {\imath} \times 2 = 2 0 \hat {\imath}$
$\vec {L} _ {0} = (0. 2) [ (1 0 \hat {\imath} - \hat {j}) \times 2 0 \hat {\imath} ] = 4 \hat {k}$
$\mathrm{AtpointA(0,-1)}$

$\vec {\tau} _ {0} = \vec {r} _ {A} \times \vec {F} = (- \hat {j}) \times 2 \hat {\imath} = 2 \hat {k}$

Ans. (49)
$L = \frac {M a ^ {2}}{3} \Omega + M \left(\frac {3 a}{4}\right) ^ {2} \Omega + \frac {M \left(\frac {a}{4}\right) ^ {2} 4 \Omega}{2}$
$L = \frac {4 9}{4 8} M a ^ {2} \Omega$
n = 4 9

Ans. (ABD)
$R _ {y} + \frac {T}{\sqrt {2}} = W + \alpha W$
... (i)
$R _ {x} = \frac {T}{\sqrt {2}}\tag{... (ii}$
Taking torque about 'O'

$W \frac {\ell}{2} + \alpha W \ell = \frac {T}{\sqrt {2}} \ell$
$T = \sqrt {2} \left(\frac {W}{2} + \alpha W\right)$
... (iii)
$R _ {x} = \frac {T}{\sqrt {2}} = \left(\frac {W}{2} + \alpha W\right)$
Taking torque about P
$R _ {\mathrm{y}} \ell = W \frac {\ell}{2}$
$R _ {y} = \frac {W}{2}$
When $T = T_{max}$
$2 \sqrt {2} W = \sqrt {2} \left(\frac {W}{2} + \alpha W\right)$
We get $\alpha=\frac{3}{2}$

Ans. (0.52)
At $t = 0,\omega = 0$
At
$t = \sqrt {\pi}, \omega = \alpha t = \frac {2}{3} \sqrt {\pi}, v = \omega r = \frac {2}{3} \sqrt {\pi}$
$\theta = \frac {1}{2} \alpha t ^ {2}$
$\theta = \frac {1}{2} \times \frac {2}{3} \times \pi = \frac {\pi}{3}$
$\theta = 6 0 ^ {\circ}$
$v _ {y} = v \sin 6 0 = \frac {\sqrt {3}}{2} V$
$h = \frac {u _ {y} ^ {2}}{2 g} = \frac {\frac {3}{4} v ^ {2}}{2 g}$
$h = \frac {\frac {3}{4} \times \frac {4}{9} \pi}{2 g}$
$h = \frac {3 \pi}{9 \times 2 g} = \frac {\pi}{6 g}$
Maximum height from plane,
$H = \frac {R}{2} + h$
$H = \frac {1}{2} + \frac {\pi}{6 \times 1 0}$
$x = \frac {\pi}{6}; x = 0. 5 2$

Ans. (2.85 to 2.86)
Solid sphere 1kg, 1m
5 + 1 - 1 - f = 1a
5 - f = a
About COM $f1 - 2(1(0.5)) = \frac{2}{5}Mr^{2}\alpha$ $\Rightarrow \quad f - 1 = \frac{2}{5}a$ $\Rightarrow \quad f = 1 + \frac{2}{5}a$ $5 - a = 1 + \frac{2}{5}a$ $\Rightarrow \quad 4 = \frac{7a}{5}$ $\Rightarrow \quad a = \frac{20}{7} = 2.86 m/s^{2}$

Ans. (3)
Torque about origin is zero
So angular momentum about origin remains conserved.
$\left| \begin{array}{c c c} \hat {i} & \hat {j} & \hat {k} \\ \frac {1}{\sqrt {2}} & \sqrt {2} & 0 \\ - \sqrt {2} & \sqrt {2} & \frac {2}{\pi} \end{array} \right| = \left| \begin{array}{c c c} \hat {i} & \hat {j} & \hat {k} \\ x & y & 0. 5 \\ v _ {x} & v _ {y} & \frac {2}{\pi} \end{array} \right|$
$\hat {\imath} \left[ \sqrt {2} \times \frac {2}{\pi} \right] - \hat {\jmath} \left[ \frac {\sqrt {2}}{\pi} \right] + \hat {k} [ 1 + 2 ]$
$= \hat {\imath} \left[ \frac {y \times 2}{\pi} - 0. 5 v _ {y} \right] - \hat {\jmath} \left[ \frac {x \times 2}{\pi} - 0. 5 v _ {x} \right] + \hat {k} \big [ x v _ {y} - y v _ {x} \big ]$
$x _ {v y} - y _ {v x} = 3$

Ans. (B)
$v = \omega_{0}R$ ; no slipping
$\begin{array}{r l} \therefore & \omega_ {0} = 2 \omega \\ & \vec {L} = m \vec {r} \times \vec {v} _ {c} + I _ {c} \omega_ {0} \\ & = M 2 R v + \frac {1}{2} M R ^ {2} \omega_ {0} \\ & = 4 M R ^ {2} \omega + \frac {1}{2} M R ^ {2} (2 \omega) = 5 M R ^ {2} \omega \end{array}$

Ans. (A)
$\begin{array}{c} \text { M,L } \quad \omega \\ \text { v } \\ \text { u } \\ \text { m } \end{array}$
Applying angular momentum conservation about hinge
$m v \frac {L}{2} + 0 = - m v \frac {L}{2} + \frac {M L ^ {2}}{3} \omega\tag{... (i}$
Also, from equation of restitution
$e = 1 = \frac {\omega \frac {L}{2} + V}{u} \Rightarrow u = \omega \frac {L}{2} + V\tag{... (ii}$
Solving (i) and (ii)
$\omega \approx 6.98$ rad/sec and $v = 4.30m / s$
Hence option (A)

Ans. (10)
$T = \frac {2 \pi}{\omega} = 2 \pi \sqrt {\frac {I}{C}}$
$\Rightarrow \omega = \sqrt {\frac {C}{I}}$
Where I = moment of inertia
$I = (3 0) (4) ^ {2} + (2 0) (6) ^ {2}$
$= 1 2 0 0 g m \cdot c m ^ {2}$

$= 1. 2 \times 1 0 ^ {- 4} k g \cdot m ^ {2}$
$\Rightarrow \omega = \sqrt {\frac {1 . 2 \times 1 0 ^ {- 8}}{1 . 2 \times 1 0 ^ {- 4}}}$
$\Rightarrow \omega = \sqrt {1 0 ^ {- 4}}$
$\omega = (1 0 ^ {- 2})$
$n \times 1 0 ^ {- 3} = 1 0 ^ {- 2} \Rightarrow n = 1 0$

Ans. (ABCD)
$J _ {0} = m v$
... (1)
$J _ {0} h _ {m} = I _ {\mathrm{c}} \omega$
...(2)
$v = \omega R$
... (3)
$\Rightarrow h _ {m} = \frac {I _ {C}}{m R}$
(A) If $a = 0$ , $I_{c} = \frac{1}{2} mb^{2}$ and $R = b$ $\therefore h_m = \frac{b}{2}$

(B) If $a = b$ , $I_C = mb^2$ and $R = b \therefore h_m = b$
(C) $v = \frac{J_0}{m} \Rightarrow 100 = \frac{V}{R} = \frac{J_0}{mR}$
(D) Force is acting on COM ∴ No rotation.

Ans. (30)
J = m v
$J r = I _ {\mathrm{c}} \omega$
... (1)
$J _ {C} = \frac {1}{4} m R ^ {2}$
... (2)
... (3)
$t = \frac {2 v}{g}$
... (4)
$\theta = 2 \pi N = \omega t$
... (5) $\therefore N = 30$

## JEE (Main) Practice Paper

Ans. 15
For a body
$mg \sin\theta - f = ma$ and $fr = I\alpha, a = \alpha r$
$mg \sin \theta - \frac{I\alpha}{r} = ma$
$mg \sin \theta = \frac{Ia}{r^{2}} + ma$
$a = \frac {m g \sin \theta}{\left(\frac {1}{r ^ {2}} + m\right)} \Rightarrow a _ {s c} = \frac {2}{3} g s i n \theta$
and $a_{bc} = \frac{1}{2} g\sin \theta$
Relative acceleration $= \frac{2}{3} g\sin \theta -\frac{1}{2} g\sin \theta = \frac{1}{6} g\sin \theta = \frac{g}{12}$
Gap after time t = 6
$t = 6, \frac {1}{2} \times \frac {g}{1 2} \times (6) ^ {2} = \frac {1}{2} \times \frac {1 0}{1 2} \times 3 6 = 1 5 m$

Ans. $\frac{l^2}{12x}$
$V _ {C M} = \frac {I}{m}$
$I _ {x} = \frac {m \ell^ {2}}{1 2} \omega$
$\mathbf {v} _ {c m} - w y = 0$

$\Rightarrow y = \frac {v _ {c m}}{\omega}$

Ans. (D)
For the semicircular lamina of mass m, the moment of inertia about an axis through C is $I_{c} = \frac{1}{2} mr^{2}$ Let $I_{CM} = \text{moment of inertia about an axis through its centre of mass}$ .
$I _ {\mathrm{C}} = I _ {\mathrm{CM}} + m d ^ {2} = \frac {1}{2} m r ^ {2} - m \left(\frac {4 r}{3 \pi}\right) ^ {2} = \left(\frac {1}{2} - \frac {1 6}{9 \pi^ {2}}\right) m r ^ {2}$

Ans. $\frac{mv}{4}$ Angular momentum conservation about O.
∴ $mV\frac{\ell}{2}=0+\frac{m\ell^{2}}{3}\omega$ ∴ $\omega\ell=V\frac{3}{2}$ Now $V_{CM}$ of rod = $\frac{\omega\ell}{2} = \frac{3V}{4}$ ∴ $I_{1}$ on (1) = mv $I_{1}-N_{imp}=m\frac{3v}{4}$ ∴ $mv-N_{Im}=\frac{3mv}{4}$ ∴ $N_{Im}=\frac{mv}{4}$

Ans. $\frac{2mv_0^2}{l}$ $(V_{cm}$ of rod $= 0$ as $P_{i} = P_{f})$ $L_{i} = L_{f}$ $\therefore \omega = \frac{2v_0}{\ell}$ $T = m\omega^2\frac{\ell}{2} = \frac{m4v_0^2}{\ell^2} \times \frac{\ell}{2} = \frac{2mv_0^2}{\ell}$