JEE Physics Circular Motion MCQs Set 13

Answer: (d) (ii) and (iii) only

Explanation:
1. Friction must supply both the centripetal force \( \frac{Mv^2}{r} \) and the tangential force \( Ma \).
2. So its magnitude is \( M\sqrt{\frac{v^4}{r^2}+a^2} \), greater than \( \frac{Mv^2}{r} \), and (i) is false because there is a tangential component.
3. Also \( \mu Mg\ge Ma \) so \( \mu\ge\frac ag \): (iii) is true.

Teacher's Note:
1. When the speed changes, the net acceleration has a tangential part and does not point to the centre.
2. Friction is the vector sum of two perpendicular requirements.

Answer: (a) (ii), (iii) and (iv) only

Explanation:
1. Since \( \frac{mv^2}{r}=mg \): at A, \( N-mg=mg\Rightarrow N=2w \); at E, \( mg-N=mg\Rightarrow N=0 \).
2. At G and C the centripetal acceleration is horizontal, so the vertical force balance gives N = w.
3. So A exceeds E by 2w, G reads w, and E reads zero. The readings are not all equal.

Teacher's Note:
1. Apply Newton's law along the radius at each point, remembering that the cage stays upright.
2. The reading is largest at the bottom and zero at the top for this particular speed.

Answer: (b) (i) and (iv) only

Explanation:
1. After a year the displacement is zero, so the average velocity is zero (i true) but the distance covered is not zero, so the average speed is not zero (ii false).
2. In six months the velocity reverses, so \( \Delta v=2v \) and the average acceleration is not zero (iii false).
3. The centripetal acceleration points to the Sun (iv true).

Teacher's Note:
1. Average velocity uses displacement, average speed uses path length.
2. Half an orbit reverses the velocity vector.

Answer: (c) (ii) and (iii) only

Explanation:
1. The bank is designed for 36 km/h, so at that speed no friction is needed (i false).
2. Below the design speed the horizontal component of N is more than needed and the car slides down (ii true).
3. The road force is \( N=\sqrt{(mg)^2+(mv^2/r)^2} \), which exceeds both mg and \( mv^2/r \) (iii true).

Teacher's Note:
1. At the correct speed the resultant of weight and the normal force is exactly horizontal.
2. The normal force is the hypotenuse of the force triangle.

Answer: (d) \( h=0.1 \) m and \( v=\sqrt2 \) m/s

Explanation:
1. At the departure point \( mg\cos\theta=\frac{mv^2}{R} \) and energy gives \( v^2=2gR(1-\cos\theta) \).
2. \( \cos\theta=\frac23 \), so \( h=R(1-\cos\theta)=0.1 \) m.
3. \( v^2=2\times10\times0.1=2 \), so \( v=\sqrt2 \) m/s.

Teacher's Note:
1. Contact is lost when the normal reaction becomes zero.
2. Combine that condition with energy conservation.

Answer: (a) (ii), (iii) and (iv) only

Explanation:
1. \( \frac{mv^2}{r}\propto t^2\Rightarrow v\propto t \), so the speed grows uniformly and the tangential force is constant (i false).
2. Power \( =F_tv\propto t \) and the average power over t seconds is also proportional to t (ii, iii true).
3. \( \tan\phi=\frac{F_t}{F_r}\propto\frac{1}{t^2} \) decreases (iv true).

Teacher's Note:
1. Speed proportional to time means constant tangential acceleration.
2. Only the tangential force does work, and its power is force times speed.

Answer: (b) -8

Explanation:
1. For circular motion about the centre, \( \vec\omega\perp\vec v \).
2. \( \vec v\cdot\vec\omega=3x+24=0\Rightarrow x=-8 \).

Teacher's Note:
1. The angular velocity vector is perpendicular to the plane of the circle, hence to the velocity.
2. Set the dot product to zero.

Answer: (d) \( -50\hat k \) m/s²

Explanation:
1. \( \vec a=\vec\omega\times\vec v=(-8\hat i+6\hat j)\times(3\hat i+4\hat j) \).
2. \( =(-8)(4)\hat k+(6)(3)(-\hat k)=-32\hat k-18\hat k=-50\hat k \) m/s².

Teacher's Note:
1. For uniform circular motion the acceleration is omega cross v, directed to the centre.
2. Check magnitude: v omega = 5 x 10 = 50.

Answer: (a) 36 rad/s

Explanation:
1. \( T=m\omega^2L\Rightarrow\omega^2=\frac{324}{0.5\times0.5}=1296 \).
2. \( \omega=36 \) rad/s.

Teacher's Note:
1. If the string stays essentially horizontal, tension supplies the centripetal force.
2. Take the square root of 1296.

Answer: (b) 14 m/s²

Explanation:
1. \( v=\frac{ds}{dt}=3t^2=12 \) m/s, \( a_t=6t=12 \) m/s², \( a_r=\frac{144}{20}=7.2 \) m/s².
2. \( a=\sqrt{12^2+7.2^2}=\sqrt{195.8}\approx14 \) m/s².

Teacher's Note:
1. Find both components of acceleration at the given instant.
2. Do not report either component alone.

Answer: (c) \( -\frac{v^2}{R}\cos\theta\,\hat i-\frac{v^2}{R}\sin\theta\,\hat j \)

Explanation:
1. The acceleration has magnitude \( \frac{v^2}{R} \) and points from P to the centre, opposite to the position vector \( R(\cos\theta\,\hat i+\sin\theta\,\hat j) \).
2. Hence \( \vec a=-\frac{v^2}{R}(\cos\theta\,\hat i+\sin\theta\,\hat j) \).

Teacher's Note:
1. Centripetal acceleration is anti-parallel to the position vector from the centre.
2. The tangential direction would be perpendicular to this.

Answer: (d) \( T\propto R^{(n+1)/2} \)

Explanation:
1. Force \( F\propto R^{-n}=m\omega^2R \) so \( \omega^2\propto R^{-(n+1)} \).
2. \( T=\frac{2\pi}{\omega}\propto R^{(n+1)/2} \).

Teacher's Note:
1. The central force supplies the centripetal force m omega^2 R.
2. For n = 2 this reproduces Kepler's law T squared proportional to R cubed.

Answer: (a) zero

Explanation:
1. Force magnitude \( F=-\frac{dU}{dr}=-\frac{k}{r^3} \) (attractive). Centripetal: \( \frac{mv^2}{a}=\frac{k}{a^3}\Rightarrow mv^2=\frac{k}{a^2} \).
2. KE \( =\frac{k}{2a^2} \) and PE \( =-\frac{k}{2a^2} \), so total = 0.

Teacher's Note:
1. Use force = minus the gradient of the potential, then apply the circular-orbit condition.
2. KE and PE are equal and opposite here.

Answer: (b) \( \frac{m\omega^2}{k} \)

Explanation:
1. Extension x: \( kx=m\omega^2(l+x)\Rightarrow x=\frac{m\omega^2l}{k-m\omega^2} \).
2. For \( k\gg m\omega^2 \), \( \frac{x}{l}\approx\frac{m\omega^2}{k} \).

Teacher's Note:
1. The spring force supplies the centripetal force at the stretched radius l + x.
2. Neglect m omega^2 compared with k in the denominator.

Answer: (c) 0.34 %

Explanation:
1. Reading \( =mg-m\omega^2R \), so the fractional reduction is \( \frac{\omega^2R}{g} \).
2. \( \omega=\frac{2\pi}{86400}=7.27\times10^{-5} \) rad/s, so \( \frac{\omega^2R}{g}=\frac{(7.27\times10^{-5})^2\times6.4\times10^6}{9.8}\approx3.4\times10^{-3} \).

Teacher's Note:
1. The reduction in weight is the centripetal term m omega^2 R.
2. Convert the day to seconds first.

Answer: (d) 2.0 hours

Explanation:
1. Half weight: \( m\omega^2R=\frac{mg}{2}\Rightarrow\omega=\sqrt{\frac{g}{2R}} \).
2. \( T=2\pi\sqrt{\frac{2R}{g}}=2\pi\sqrt{\frac{2\times6.4\times10^6}{9.8}}\approx7.2\times10^3 \) s, about 2.0 hours.

Teacher's Note:
1. Set the centripetal term equal to half of mg.
2. Then convert the period from seconds to hours.

Answer: (a) 975 N and 1025 N

Explanation:
1. \( v=5 \) m/s so \( \frac{mv^2}{r}=\frac{100\times25}{100}=25 \) N.
2. At B (top of a hump): \( mg-N=25\Rightarrow N=975 \) N.
3. At D (bottom of a dip): \( N-mg=25\Rightarrow N=1025 \) N.

Teacher's Note:
1. On a convex arc the centre of curvature is below, so N is less than mg.
2. On a concave arc the centre is above, so N is greater than mg.

Answer: (b) \( \frac{50}{3} \) s

Explanation:
1. Belt speed is common: \( \omega_Ar_A=\omega_Br_B \).
2. \( \omega_B=100\times\frac{2\pi}{60}=\frac{10\pi}{3} \) rad/s, so \( \omega_A=\frac{10\pi}{3}\times\frac{25}{10}=\frac{25\pi}{3} \) rad/s.
3. \( t=\frac{\omega_A}{\alpha_A}=\frac{25\pi/3}{\pi/2}=\frac{50}{3} \) s.

Teacher's Note:
1. Points on the belt, and on the rims of both wheels, have the same linear speed.
2. The smaller wheel must spin faster by the ratio of radii.

Answer: (c) \( \frac{5v}{4} \) and \( \frac{5v}{4R} \)

Explanation:
1. P stays on the circle: its coordinate perpendicular to CD is \( x=R\sin\phi \). Differentiating, \( v=R\cos\phi\,\dot\phi \).
2. With \( \cos\phi=\frac45 \): \( \dot\phi=\frac{v}{R\cos\phi}=\frac{5v}{4R} \).
3. Speed of P \( =R\dot\phi=\frac{5v}{4} \).

Teacher's Note:
1. The point P moves on the circle, so its speed is R times its angular speed.
2. The rod's speed gives the rate of change of the coordinate x of P.