Work, Power & Energy

Ans. (A-P,Q), (B-P,Q,S), (C-P,Q,R), (D-P,Q,R)
No slipping anywhere.
Net force is centripetal as v=constant.

Ans. (A-R), (B-S) (C-Q), (D-P)
(A) $a_{t}=0, a_{C}\neq0$ (B) $a_{t}\neq0, a_{C}\neq0$ (C) $a_{t}\neq0, a_{C}\neq0$ (D) $a_{t}\neq0, a_{C}=0 (v=0)$

Ans. (A-Q), (B-P), (C-Q), (D-S) $y_{A}=2y_{B}\Rightarrow v_{A}=2v_{B}$ A moves $\downarrow$ and
B moves $\uparrow$ as
Tension on B is more than A. $mgh-\frac{mgh}{2}=\frac{1}{2}mv_{A}^{2}+\frac{1}{2}mv_{B}^{2}$ $\frac{gh}{2}=\frac{5}{2}v_{B}^{2}$ $\Rightarrow v_{B}=\sqrt{\frac{5}{5}}=1, v_{A}=2$ mg-T=maA
2T-mg=maB $a_{A}=2a_{B}$

Ans. (A-Q), (B-P), (C-Q), (D-P)
A) $W_{N}+W_{mg}=\Delta KE$ $V_{f}=at=5\times2=10\ m/s$ $W_{N}=\frac{1}{2}mV_{f}^{2}-0=\frac{1}{2}\times\frac{1}{2}\times10^{2}=25J$
B) $W_{mg}=0$ (No horizontal displacement)
C) $W_{net}=\Delta KE=25J$
D) $W.D_{out}=0(S=0)$

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. .. (