Circular Motion

Answer: (d)

1. One revolution is an angular displacement of $2\pi$ rad.
2. $\theta = 2 \times 2\pi = 4\pi$ rad (the radius does not matter).
3. So the answer is (d) $4\pi$.

Answer: (b)

1. In half a period the particle goes to the diametrically opposite point.
2. Displacement = diameter $= 2R$.
3. Distance = half the circumference $= \pi R$.
4. So the answer is (b) $2R, \pi R$.

Answer: (c)

1. Angular speed $\omega = v/r$.
2. Angle in $t = 2$ s: $\theta = \omega t = 2v/r$.
3. So the answer is (c) $2v/r$.

Answer: (a)

1. $\omega = 2\pi/T$ depends only on the period, not on mass or radius.
2. Same $T \Rightarrow \omega_1 = \omega_2$.
3. So the answer is (a) $1:1$.

Answer: (c)

1. From one end of a diameter to the other is half a revolution: $\theta = \pi$ rad.
2. $\omega = \theta/t = \pi/3$ rad/s.
3. So the answer is (c) $\frac{\pi}{3}$.

Answer: (c)

1. Hour hand: $T = 12$ h $= 43200$ s.
2. $\omega = \frac{2\pi}{43200} = \frac{\pi}{21600}$ rad/s.
3. So the answer is (c) $\frac{\pi}{21600}$.

Answer: (b)

1. Second hand: $\omega_s = \frac{2\pi}{60} = \frac{60\pi}{1800}$ rad/s.
2. Minute hand: $\omega_m = \frac{2\pi}{3600} = \frac{\pi}{1800}$ rad/s.
3. $\omega_s - \omega_m = \frac{59\pi}{1800}$ rad/s.
4. So the answer is (b) $\frac{59\pi}{1800}$.

Answer: (c)

1. Minute hand: $\omega_m = \frac{2\pi}{3600} = \frac{12\pi}{21600}$ rad/s.
2. Hour hand: $\omega_h = \frac{2\pi}{43200} = \frac{\pi}{21600}$ rad/s.
3. $\omega_m - \omega_h = \frac{11\pi}{21600}$ rad/s.
4. So the answer is (c) $\frac{11\pi}{21600}$.

Answer: (b)

1. The minute hand turns $360^\circ$ in $3600$ s.
2. $\omega = 360/3600 = 0.1^\circ$/s.
3. So the answer is (b) $0.1$.

Answer: (b)

1. $\omega = \frac{2\pi x}{t}$ for $x$ revolutions in time $t$.
2. $v = r\omega = \frac{\pi}{2} \cdot \frac{2\pi x}{t} = \frac{\pi^2 x}{t}$.
3. So the answer is (b) $\frac{\pi^2 x}{t}$.

Answer: (d)

1. $\omega = 2\pi/T \Rightarrow \omega \propto 1/T$.
2. $\frac{\omega_B}{\omega_A} = \frac{T_A}{T_B} = 3$.
3. So the answer is (d) $3:1$.

Answer: (c)

1. $\vec v = \frac{d\vec r}{dt} = -a\omega\sin\omega t\,\hat i + a\omega\cos\omega t\,\hat j$.
2. $\vec r \cdot \vec v = -a^2\omega\sin\omega t\cos\omega t + a^2\omega\sin\omega t\cos\omega t = 0$.
3. Zero dot product means $\vec r \perp \vec v$.
4. So the answer is (c) $90^\circ$.

Answer: (c)

1. A and B are ends of a diameter. Path $A\to B \to A \to B$ is three half-circles.
2. Distance $= 3 \times \pi r = 3\pi r$.
3. The runner ends at B, so displacement $= AB = 2r$.
4. So the answer is (c) $3\pi r, 2r$.

Answer: (c)

1. $\omega = \frac{2\pi x}{t}$ for $x$ revolutions in time $t$.
2. $v = r\omega = \frac{\pi}{2} \cdot \frac{2\pi x}{t} = \frac{\pi^2 x}{t}$.
3. So the answer is (c) $\frac{\pi^2 x}{t}$.

Answer: (a)

1. Going $P \to Q \to R \to S$ covers $3/4$ of the circle, but only the end points matter for displacement.
2. $OP \perp OS$, so $PS = \sqrt{r^2 + r^2} = r\sqrt{2}$.
3. $PS = 2\sqrt{2} = \sqrt{8}$ km.
4. So the answer is (a) $\sqrt{8}$ km.

Answer: (b)

1. Reaching the other end of the diameter means half a revolution: $\theta = \pi$ rad.
2. $\omega = \theta/t = \pi/3$ rad/s.
3. So the answer is (b) $\frac{\pi}{3}$.

Answer: (d)

1. Same period $\Rightarrow$ same angular speed $\omega$.
2. $v = r\omega \Rightarrow v \propto r$.
3. $v_2 : v_1 = r_2 : r_1$.
4. So the answer is (d) $r_2 : r_1$.

Answer: (c)

1. In 30 min the second hand makes 30 revolutions: $s_1 = 30 \times 2\pi(0.75) = 45\pi$ m.
2. The minute hand makes half a revolution: $s_2 = \pi(0.60) = 0.6\pi$ m.
3. $x = s_1 - s_2 = 44.4\pi = 44.4 \times 3.14 \approx 139.4$ m.
4. So the answer is (c) 139.4.

Answer: (c)

1. At latitude $\lambda$ the radius of the circle is $R\cos\lambda$, and $\omega$ is the same.
2. $v = \omega R\cos\lambda = V\cos 30^\circ = \frac{\sqrt3}{2}V$.
3. So the answer is (c) $\frac{\sqrt{3}}{2}V$.

Answer: (c)

1. Rolling through $180^\circ$ moves the centre forward by $\pi R = \pi$ m.
2. The contact point goes from the bottom to the top: it rises $2R = 2$ m.
3. Displacement $= \sqrt{\pi^2 + 2^2} = \sqrt{\pi^2 + 4}$ m.
4. So the answer is (c) $\sqrt{\pi^2 + 4}$.

Answer: (a)

1. $\omega = 2\pi/T$ depends only on the period.
2. Same $T \Rightarrow \omega_1 : \omega_2 = 1:1$.
3. So the answer is (a) 1 : 1.

Answer: (d)

1. $\omega = \frac{2\pi x}{t}$ for $x$ revolutions in time $t$.
2. $v = r\omega = \frac{\pi}{2} \cdot \frac{2\pi x}{t} = \frac{\pi^2 x}{t}$.
3. So the answer is (d) $\frac{\pi^2 x}{t}$.

Answer: (c)

1. In 3 s the body covers $3/4$ of a revolution, i.e. $270^\circ$.
2. Start and end radii are perpendicular, so displacement $= \sqrt{r^2 + r^2} = r\sqrt2$.
3. $= 10\sqrt{2}$ m.
4. So the answer is (c) $10\sqrt{2}$.

Answer: (c)

1. After half a circle the velocity reverses: $|\Delta \vec v| = 2V$.
2. Time taken $t = \frac{\pi r}{V}$.
3. $a_{avg} = \frac{2V}{\pi r / V} = \frac{2V^2}{\pi r}$.
4. So the answer is (c) $\frac{2V^2}{\pi r}$.

Answer: (b)

1. At the diametrically opposite point the velocity is reversed: $\vec v \to -\vec v$.
2. $|\Delta \vec p| = mv - (-mv) = 2mv$.
3. Speed is unchanged, so kinetic energy does not change.
4. So the answer is (b) the momentum changes by $2mv$.

Answer: (b)

1. Displacement $AB$ = chord for $120^\circ$: $2r\sin 60^\circ = r\sqrt3$.
2. Arc $AB = \frac{2\pi r}{3}$, so time $t = \frac{2\pi r/3}{\pi} = \frac{2r}{3}$ s.
3. $v_{avg} = \frac{r\sqrt3}{2r/3} = 1.5\sqrt3$ m/s.
4. So the answer is (b) $1.5\sqrt{3}$ m/s.

Answer: (d)

1. For a $90^\circ$ turn: displacement $= r\sqrt2$, arc $= \frac{\pi r}{2}$, time $t = \frac{\pi r}{2v}$.
2. $v_{avg} = \frac{r\sqrt2}{\pi r/2v} = \frac{2\sqrt2\,v}{\pi}$.
3. $\frac{v}{v_{avg}} = \frac{\pi}{2\sqrt2} = \pi : x\sqrt2 \Rightarrow x = 2$.
4. So the answer is (d) 2.

Answer: (b)

1. Arc length $s = r\theta$ with $\theta = 135^\circ = \frac{3\pi}{4}$: $r = \frac{60 \times 4}{3\pi} \approx 25.5$ m.
2. Displacement $AB = \sqrt{r^2 + r^2 - 2r^2\cos 135^\circ} = r\sqrt{2(1 + 0.7)} = r\sqrt{3.4}$.
3. $AB \approx 25.5 \times 1.84 \approx 47$ m.
4. So the answer is (b) 47 m.

Answer: 3

1. Path: $OP = r$, arc $PQ$ = quarter circle $= \frac{\pi r}{2}$, $QO = r$.
2. Distance $= 2r + \frac{\pi r}{2} = 400 + 314 = 714$ m.
3. Time $= 3$ min $58$ s $= 238$ s.
4. Average speed $= 714/238 = 3$ m/s.

Answer: (d)

1. $\omega = \frac{d\theta}{dt} = 5 \cdot \frac{\pi}{6}\cos\frac{\pi t}{6}$.
2. At $t = 3$ s: $\cos\frac{\pi}{2} = 0$, so $\omega = 0$.
3. So the answer is (d) zero.

Answer: (d)

1. $\omega \propto 1/T$. Hour hand $T_1 = 12$ h, Earth spin $T_2 = 24$ h.
2. $\omega_1 : \omega_2 = T_2 : T_1 = 24 : 12 = 2:1$.
3. So the answer is (d) 2 : 1.

Answer: (b)

1. $\theta = 4'' = \frac{4}{3600} \times \frac{\pi}{180} \approx 1.94 \times 10^{-5}$ rad.
2. $r = \frac{s}{\theta} = \frac{4.4 \times 9.46 \times 10^{15}}{1.94 \times 10^{-5}} \approx 2.15 \times 10^{21}$ m.
3. Distance for 4 revolutions $= 4 \times 2\pi r \approx 5.4 \times 10^{22}$ m.
4. Speed $= 8 \times 1.5 \times 10^{11} = 1.2 \times 10^{12}$ m/s, so $t = \frac{5.4 \times 10^{22}}{1.2 \times 10^{12}} \approx 4.5 \times 10^{10}$ s.
5. So the answer is (b) $4.5 \times 10^{10}$ s.

Answer: (c)

1. After half a revolution the velocity reverses: $|\Delta \vec v| = 2v$.
2. Time taken $t = \frac{\pi R}{v}$.
3. $a_{avg} = \frac{2v}{\pi R / v} = \frac{2v^2}{\pi R}$.
4. So the answer is (c) $\frac{2v^2}{\pi R}$.

Answer: (b)

1. Speed is constant, only direction changes by $\theta = 60^\circ$.
2. $|\Delta \vec v| = 2v\sin\frac{\theta}{2} = 2 \times 10 \times \sin 30^\circ$.
3. $|\Delta \vec v| = 10$ m/s (the vector triangle is equilateral).
4. So the answer is (b) 10 m/s.

Answer: (b)

1. As $\delta t \to 0$, $\delta \vec v$ points towards the centre (it gives the centripetal acceleration).
2. The velocity is along the tangent, which is perpendicular to the radius.
3. So the answer is (b) $90^\circ$.

Answer: (a)

1. Centripetal force $F = m\omega^2 r = \frac{4\pi^2 m r}{T^2}$.
2. $\sqrt{F} = \frac{2\pi\sqrt{mr}}{T}$
3. So the answer is (a) $\frac{2\pi\sqrt{mr}}{T}$.

Answer: (c)

1. Equal forces: $\frac{m v_1^2}{r} = \frac{3m v_2^2}{r/3} = \frac{9 m v_2^2}{r}$.
2. $v_1^2 = 9 v_2^2 \Rightarrow v_1 = 3 v_2$
3. So $n = 3$. The answer is (c) $3$.

Answer: (a)

1. Same period $\Rightarrow$ same $\omega$. Centripetal acceleration $a = \omega^2 r$.
2. $a_1 : a_2 = R : r$ (mass does not matter).
3. So the answer is (a) $R/r$.

Answer: (b)

1. (a) True: velocity is always along the tangent.
2. (b) False: in uniform circular motion the acceleration is centripetal, along the radius, not along the tangent.
3. (c) True: radial acceleration is perpendicular to tangential velocity.
4. (d) True: acceleration $\frac{v^2}{r}$ points to the centre.
5. So the answer is (b).

Answer: (b)

1. Equal forces: $\frac{m v_1^2}{r} = \frac{3m v_2^2}{r/3} = \frac{9 m v_2^2}{r}$.
2. $v_1^2 = 9 v_2^2 \Rightarrow v_1 = 3 v_2$
3. So $n = 3$. The answer is (b) 3.

Answer: (b)

1. M stays at rest, so string tension $T = Mg$.
2. The same tension gives the centripetal force on m: $Mg = m\omega^2 L$.
3. $\omega = \sqrt{\frac{Mg}{mL}}$, and $f = \frac{\omega}{2\pi}$.
4. So the answer is (b) $\frac{1}{2\pi}\sqrt{\frac{Mg}{mL}}$.

Answer: (d)

1. Spring force gives the centripetal force: $kx = m\omega^2(l + x)$, where $l$ is the natural length.
2. $8x = 0.1 \times 64\,(l + x) = 6.4(l + x)$
3. $1.6x = 6.4l \Rightarrow \frac{x}{l} = 4$
4. So the answer is (d) $4 : 1$.

Answer: 1

1. The end wall must supply the centripetal force for the whole liquid column.
2. Element $dm = \frac{2M}{L}dx$ at distance $x$ needs $dm\,\omega^2 x$.
3. $F = \int_0^L \frac{2M}{L}\omega^2 x\,dx = M\omega^2 L = M\omega^2$ (with $L = 1$ m).
4. $\omega = \sqrt{\frac{F}{M}} \Rightarrow \alpha = 1$
5. Answer: $\alpha = 1$.

Answer: (b)

1. $a = \omega^2 R = 4\pi^2 n^2 R$.
2. $\frac{a}{R} = 4\pi^2 n^2 \propto n^2$, so $x = 2$.
3. So the answer is (b) 2.

Answer: (d)

1. $r = 0.5$ m, $\omega = 2\pi n = 8\pi$ rad/s.
2. $a = \omega^2 r = 64\pi^2 \times 0.5 = 32\pi^2$ m/s$^2$.
3. So the answer is (d) $32\pi^2$.

Answer: (b)

1. Frequency $n = \frac{120}{180\text{ s}} = \frac{2}{3}\text{ Hz}$, so $\omega = 2\pi n = \frac{4\pi}{3}\text{ rad/s}$.
2. The monkey moves with the man on the same circle, so $a = \omega^2 r$.
3. $a = \frac{16\pi^2}{9} \times 9 = 16\pi^2\text{ m/s}^2$
4. So the answer is (b) $16\pi^2\text{ m s}^{-2}$.

Answer: (c)

1. Velocity and acceleration change direction every instant, so they are not constant vectors.
2. Displacement keeps changing as the particle moves.
3. Speed is constant, so $KE = \frac{1}{2}mv^2$ is constant.
4. So the answer is (c) kinetic energy.

Answer: (b)

1. In UCM velocity is along the tangent and its magnitude (speed) stays the same.
2. Acceleration is radial (towards the centre), not tangential and not along the axis.
3. So the answer is (b).

Answer: (c)

1. $F = \frac{mv^2}{r}$ with same $m$ and same $F$ gives $v^2 \propto r$.
2. $\frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$
3. So the answer is (c) $\sqrt{3} : 2$.

Answer: (a)

1. Angular momentum $L = mvr \Rightarrow v = \frac{L}{mr}$.
2. $KE = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{L^2}{m^2r^2}$
3. $KE = \frac{L^2}{2mr^2}$
4. So the answer is (a).

Answer: (b)

1. $F = \frac{mv^2}{r}$
2. $F' = \frac{(1.2m)(1.2v)^2}{1.2r} = 1.44\,F$
3. Increase $= 44\%$
4. So the answer is (b) $44\%$.