Work, Power & Energy

Ans. (A-Q,S,T), (B-Q,S,T), (C-P), (D-R,T)
For A) $v_{A}=\sqrt{2gh}=\sqrt{5gR}$ circle will be completed and $N_{A}-N_{B}=6mg$
$N _ {C} = \frac {m v _ {c} ^ {2}}{R} = 3 m g$
For B) $v_{A} = \sqrt{9gR}$ $N_{C} = \frac{mv_{c}^{2}}{R} = 7mg$
$\mathbf{For}(\mathbf{C})v_{A} = \sqrt{2gR}$ $v_{C} = 0,N_{C} = 0,F_{net}$ at $C = mg$
For D) $v_{A} = \sqrt{4gR}$ Leaves the circle in region B-C
$N _ {C} = \frac {m v _ {c} ^ {2}}{R} = 2 m g$

Ans. (A-P,Q,R), (B-Q), (C-P,Q), (D-Q,S)
A)

$v _ {0} = \sqrt {4 g R}$
⇒ Particle will just reached at highest point and come to at rest
$\Rightarrow V_{c}=0$
P) $N_{c} = mg$ (Upward)
Q) $\vec{F} + \overrightarrow{mg} = m\overrightarrow{a_{c}} \Rightarrow (\text{at point 'A'})$ Resultant of $\vec{F}$ and $\overrightarrow{mg}$ gives centripetal acceleration.
R) As particle goes up speed decreases.
B)

Particle will reach at 'B' and $v_{B}$ become = 0. (By ME conservation)
At $B \Rightarrow v_{B} = 0$ , $T = 0\left(\frac{T = mv_{B}^{2}}{R}\right)$
Particle first speed down then speed up.
C)

$\sqrt {2 g R} < v _ {A} < \sqrt {4 g R}$
• Circle won't complete.
- Between $B$ and $C$ particle comes to at rest and then it will com back with increasing speed.
• At 'B', $T_{B}=\frac{mV_{B}^{2}}{R}$
D)

Particle will complete circular motion.
• At 'B', $N_{B}=\frac{mV_{B}^{2}}{R}$
• Speed first decrease then increase.

Ans. (A-S), (B-R), (C-P), (D-Q)
$F = m g \sin \theta + \mu_ {k} m g \cos \theta \text { and } p = \vec {F}. \vec {V}$

Ans. 24

Ans. 4
$\omega_ {0} = 1 2 \mathrm{rad/sec}$
$\Delta \theta = 8 0 \mathrm{radian}$
$\omega = 2 8 \mathrm{radian/sec}$
$\omega^ {2} = \omega_ {0} ^ {2} + 2 \alpha \Delta \theta$
$\omega = \omega_ {0} + \alpha t$
$\left(\frac {\omega_ {0} + \omega}{2}\right) t = 8 0$

Ans. $\frac{10}{\pi} ms^{-2}$
Change in velocity when particle complete the half revolution: $\Delta v = v_{f} - v_{i} = 2v$
Time taken to complete the half revolution $t = \frac{\pi R}{v}$
Average acceleration $= \frac{\Delta v}{t} = \frac{2v}{\pi R / v} = \frac{2v^2}{\pi R}$ $= \frac{2\times 5^2}{\pi\times 5} = \frac{10}{\pi} m / s^2$

Ans. (i) $2\mathrm{rad} / s^2$
ii) $12+2t$ for $t \leq 2s$ , 16 for $t \geq 2s$
iii) $a = 169.01m / s^2$
iv) 44 rad
i) $\omega_{f} = \omega_{i} + 2t$
$(2 \pi) \left(\frac {9 8 0}{6 0 \pi}\right) = 1 2 + \alpha \times 2$
$a = 2 \mathrm{rad} / s ^ {2}$
ii) $v = R\omega = R(12 + 2 \times t)$ $v = 1(12 + 2t)$ After t = 2 sec it becomes constant.
iii) At t = 0.5 sec
$v = 1 3 m / \mathrm{sec}$
$a _ {C} = \frac {v ^ {2}}{R} = 1 6 9$
$a _ {t} = R \alpha = 2$
$a _ {\mathrm{total}} = \sqrt {(1 6 9) ^ {2} + (2) ^ {2}} \cong 1 6 9$
$a _ {(t = 3)} = a _ {\mathrm{total}} = a _ {C} = \frac {v ^ {2}}{R} = \frac {1 6 ^ {2}}{1} = 2 5 6 m / s ^ {2}$
iv) Angular displacement
$\theta = \omega t + \frac {1}{2} \alpha t ^ {2}$
Putting t = 2 sec we get ans.

Ans. $\frac{u^2\cos^2\theta}{g\cos^3\left(\frac{\theta}{2}\right)}$
$R = \frac {V ^ {2}}{g \cos (\frac {\theta}{2})}$
$R = \frac {V _ {x} ^ {2} + v _ {y} ^ {2}}{g \cos (\frac {\theta}{2})} = \frac {v _ {x} ^ {2} + v _ {x} ^ {2} \tan^ {2} (\frac {\theta}{2})}{g \cos (\frac {\theta}{2})}$
$R = \frac {v _ {x} ^ {2} [ 1 + \tan^ {2} (\frac {\theta}{2}) ]}{g \cos (\frac {\theta}{2})}$
$= \frac {v _ {x} ^ {2} \sec^ {2} (\frac {\theta}{2})}{g \cos (\frac {\theta}{2})}$
$= \frac {v _ {x} ^ {2}}{g \cos^ {3} (\frac {\theta}{2})} = \frac {u ^ {2} \cos^ {2} \theta}{g \cos^ {3} (\frac {\theta}{2})}$

Ans. (i) $75m / s^2$ , (ii) $125m / s^2$ $\tan 37^\circ = \frac{a_t}{a_c}$ $\therefore \quad \frac{3}{4} = \frac{(dv / dt)}{v^2 / R}$ (i) $\frac{dv}{dt} = \left(\frac{3}{4}\right)\left(\frac{v^2}{R}\right)$ (ii) $a = \sqrt{\left(\frac{dv}{dt}\right)^2 + \left(\frac{v^2}{R}\right)^2}$

Ans. $\sqrt{2g}$ rad/s
$T = 2 m g = m r \omega^ {2}$
$\omega = \sqrt {\frac {2 g}{r}} = \sqrt {2 g} \mathrm{rad/s}.$

Ans. 5
$\mu N = m g \quad \Rightarrow \frac {\mu m v ^ {2}}{r} = m g \quad \Rightarrow v = \sqrt {\frac {g r}{\mu}} = 2 5 m / s$

Ans. 5
$T \cos 3 0 = m (L \cos 3 0) \omega^ {2}$
$T \sin 3 0 + N = m g$
$N = m g - T \sin 3$....(i)
Putting value of T from equation in (ii) we get N. .. (

Ans. $T = 6.6N, v_{\mathrm{max}} = 35ms^{-1}$
$v = r \omega = (1.5) (2 \pi) \left(\frac {4 0}{6 0}\right)$
$T = \frac {m v ^ {2}}{R} = \frac {0.25 \times 4 \pi^ {2}}{1.5} = 6.6 N$
$2 0 0 = \frac {m v _ {m a x} ^ {2}}{R}$
$v _ {\max} = 3 5 m s ^ {- 1}$

Ans. $\frac{15\pi^{2}}{10}=14.8N$ , This force is exerted by blade of fan and equal force is exerted by particle on blade in same magnitude but opposite in direction. $\omega=2\pi n=\frac{2\pi\times1500}{60}$ rad/sec $r=\frac{d}{2}=60\ cm=0.6\ m$ $m=1\ g=10^{-3}\ kg$ $F=m\omega^{2}r=10^{-3}\times\left(\frac{2\pi\times1500}{60}\right)^{2}\times0.6=\frac{15\pi^{2}}{10}=14.8$
This force is exerted by blade of fan and equal force is exerted by particle on blade in same magnitude but opposite in direction.

Ans.

$T - m \omega^ {2} R = m a....(i)$
$2 m \omega^ {2} R - T = 2 m a$
(i) + (ii)
$m w ^ {2} R = 3 m a....(ii)$
$a = \frac {\omega^ {2} R}{3}$
Put in eq (i)
$T - m \omega^ {2} R = a = \frac {\omega^ {2} R}{3}$
$T = m \omega^ {2} R + \frac {m \omega^ {2} R}{3} = \frac {4 m \omega^ {2} R}{3}$

Ans. Yes, $a_{c}=8$ , $\mu g=1$ $v=\sqrt{\mu rg}$ So, required value of $\mu=\frac{v^{2}}{rg}=\frac{25}{3\times10}=\frac{5}{6}=0.8$ $\mu=0.1$ So, cyclist will slip.

Ans. $\frac{49}{80}$ Coefficient of friction $\mu \geq \frac{v^2}{rg}$ $\Rightarrow \mu_{\min} = \frac{(4.9)^2}{4\times 9.8} = \frac{49}{80}$

Ans. $5\sqrt{2} ms^{-1}$ For safe driving, $v_{\max} = \sqrt{\mu rg}$ $10 = \sqrt{\mu rg}$ For wet road $v' = \sqrt{\frac{\mu}{2} rg} = \frac{10}{\sqrt{2}} = 5\sqrt{2} m / s$

Ans. 2.0
FBD of block in ground frame

$N\cos \theta = mg$

$\begin{array}{rl} & N\sin \theta = m\omega^2 r\quad [\mathrm{centripetalforce}]\\ \Rightarrow & \tan \theta = \omega^2 r / g\\ \Rightarrow & \omega = \sqrt{\frac{g\tan\theta}{r}} = \sqrt{\frac{g}{h}} (\tan \theta) = 2\times 1 = 2\mathrm{rad / s} \end{array}$