JEE Physics Motion In A Plane MCQs Set 10

Answer: (a) maximum at A

Explanation:
1. In a valley the normal reaction is \(N=mg+\frac{mv^2}{R}\), and on a crest it is \(N=mg-\frac{mv^2}{R}\).
2. The speed is the same everywhere, so N is greater in a valley than on the crest, and in the valleys it is larger where R is smaller.
3. The valley A has the smaller radius, so the normal reaction is maximum at A, and it is minimum on the crest B.

Teacher's Note:
1. In a valley the centripetal force points up, so the road must push harder than the weight.
2. On a crest the road pushes less than the weight.

Answer: (b) \(\sqrt{gR\left(\frac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}\right)}\)

Explanation:
1. At the maximum speed friction acts down the slope with its limiting value. Radially \(N\sin\theta+\mu_sN\cos\theta=\frac{mv^2}{R}\).
2. Vertically \(N\cos\theta-\mu_sN\sin\theta=mg\).
3. Dividing, \(\frac{v^2}{Rg}=\frac{\sin\theta+\mu_s\cos\theta}{\cos\theta-\mu_s\sin\theta}=\frac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}\), so \(v=\sqrt{gR\,\frac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}}\).

Teacher's Note:
1. Resolve forces along the radius and along the vertical, then divide the two equations.
2. Only the formula with gR inside the root has the dimensions of speed.

Answer: (c) \(\frac{\pi}{2}<\theta<\frac{3\pi}{4}\)

Explanation:
1. With a rigid rod the bob can just reach the top with zero speed, so \(\frac12Mv^2=Mg(2L)\) and \(v^2=4gL\).
2. When the speed is \(\frac{v}{2}\): \(\frac12Mv^2=\frac12M\frac{v^2}{4}+MgL(1-\cos\theta)\).
3. \(2=\frac12+1-\cos\theta\), so \(\cos\theta=-\frac12\) and \(\theta=120^\circ=\frac{2\pi}{3}\), which lies between \(\frac{\pi}{2}\) and \(\frac{3\pi}{4}\).

Teacher's Note:
1. For a rod (unlike a string) the minimum speed at the top is zero, so \(v^2=4gL\).
2. Apply conservation of energy between the lowest point and the point at angle θ.

Answer: (a) \(\sqrt{gl}\)

Explanation:
1. The bob falls through a height \(h=l(1-\cos60^\circ)=\frac{l}{2}\).
2. By conservation of energy \(\frac12mv^2=mgh\).
3. \(v=\sqrt{2g\frac{l}{2}}=\sqrt{gl}\).

Teacher's Note:
1. The height lost is measured from the lowest point, l(1 - cos θ), not l cos θ.
2. Then use energy conservation.

Answer: (b) the acceleration vector is tangent to the circle

Explanation:
1. In uniform circular motion the velocity is along the tangent, which makes (a) true.
2. The acceleration is centripetal, directed along the radius towards the centre, which makes (c) true, and the acceleration is perpendicular to the velocity, which makes (d) true.
3. So the statement that the acceleration is tangent to the circle is false.

Teacher's Note:
1. Tangential acceleration exists only when the speed changes.
2. For a constant angular speed the only acceleration is the radial one.

Answer: (c) 24.2 m

Explanation:
1. For a given banking angle \(\tan\theta=\frac{v^2}{rg}\), so \(r\propto v^2\).
2. \(\frac{r_2}{r_1}=\left(\frac{1.1v}{v}\right)^2=1.21\).
3. \(r_2=1.21\times20=24.2\) m.

Teacher's Note:
1. The radius varies with the square of the speed.
2. A 10% rise in speed needs a 21% larger radius.

Answer: (c) \(\frac23\) s

Explanation:
1. The total acceleration makes 45° with the radial direction when \(a_r=a_t=9\ \text{m/s}^2\).
2. \(\frac{v^2}{r}=9\) gives \(v^2=36\), so \(v=6\) m/s.
3. From \(v=a_tt\): \(t=\frac{6}{9}=\frac23\) s.

Teacher's Note:
1. An angle of 45° means the radial and tangential components are equal.
2. Use the tangential acceleration to find the time from the speed.

Answer: (c) \(\tan30^\circ\)

Explanation:
1. The velocity is along the tangent. The tangential acceleration is the component of acceleration along the velocity, and the centripetal acceleration is perpendicular to it.
2. If the total acceleration makes angle \(\theta\) with the velocity, \(\tan\theta=\frac{a_c}{a_t}\).
3. So \(\frac{a_c}{a_t}=\tan30^\circ\).

Teacher's Note:
1. Tangential and radial components are perpendicular, so the ratio of the perpendicular to the parallel one is the tangent of the angle.
2. The angle is measured from the velocity, that is the tangential direction.

Answer: (a) \(g\sqrt{10}\)

Explanation:
1. To just complete the circle the speed at the lowest point satisfies \(u^2=5gR\).
2. At the level of the centre, \(v^2=u^2-2gR=3gR\), so \(a_c=\frac{v^2}{R}=3g\) (towards the centre).
3. The tangential acceleration there is g (downward), so \(a=\sqrt{(3g)^2+g^2}=g\sqrt{10}\).

Teacher's Note:
1. At the level of the centre the weight is entirely tangential, so it gives a tangential acceleration g.
2. Add the radial and tangential accelerations as perpendicular vectors.

Answer: (d) \(a_r(a_t)^2\)

Explanation:
1. Let \(a_r=\frac{v^2}{r}=kt\). Then \(v^2=krt\).
2. Differentiating, \(2v\frac{dv}{dt}=kr\), so \(a_t=\frac{kr}{2v}\), which gives \(a_t^2=\frac{k^2r^2}{4v^2}=\frac{kr}{4t}\).
3. \(a_ra_t^2=kt\times\frac{kr}{4t}=\frac{k^2r}{4}\), which does not depend on t.

Teacher's Note:
1. Express both accelerations in terms of v and t and look for the combination in which t cancels.
2. The other options all change with time.

Answer: (c) \(4\pi^2\)

Explanation:
1. The angular speed is \(\omega=2\times2\pi=4\pi\) rad/s.
2. \(a=\omega^2r=(4\pi)^2\times0.25\).
3. \(a=4\pi^2\ \text{m/s}^2\).

Teacher's Note:
1. Convert revolutions per second into radians per second.
2. The acceleration in uniform circular motion is \(\omega^2r\), directed to the centre.

Answer: (b) 60°

Explanation:
1. At the lowest position \(T=mg+\frac{mv^2}{l}\), and \(v^2=2gl(1-\cos\theta)\).
2. \(T=mg(3-2\cos\theta)\).
3. \(2940=150\times9.8\times(3-2\cos\theta)\) gives \(3-2\cos\theta=2\), so \(\cos\theta=\frac12\) and \(\theta=60^\circ\).

Teacher's Note:
1. The tension is greatest at the lowest point.
2. Use the formula \(T=mg(3-2\cos\theta)\) for a pendulum released from angle θ.

Answer: (d) 6 m/s

Explanation:
1. The initial energy is \(mgl\), with \(l=2\) m.
2. At the lowest point the kinetic energy is 90% of it: \(\frac12mv^2=0.9\,mgl\).
3. \(v=\sqrt{1.8gl}=\sqrt{1.8\times10\times2}=6\) m/s.

Teacher's Note:
1. The mass cancels in this problem, so the 200 g is not needed.
2. With no loss of energy the speed would be \(\sqrt{2gl}\), which is about 6.3 m/s.

Answer: (c) R/2

Explanation:
1. The height of A above the lowest point is \(R(1-\cos53^\circ)=0.4R\) and that of B is \(R(1-\cos37^\circ)=0.2R\).
2. The speed at B is given by \(v^2=2g(0.4R-0.2R)=0.4gR\).
3. The velocity at B is along the tangent, at \(37^\circ\) to the horizontal, so the component of g perpendicular to the velocity is \(g\cos37^\circ=0.8g\).
4. The radius of curvature is \(\frac{v^2}{g\cos37^\circ}=\frac{0.4gR}{0.8g}=\frac R2\).

Teacher's Note:
1. Once the block leaves the track it is a projectile, and the centripetal acceleration is the component of g perpendicular to the velocity.
2. The radius of curvature of a path is the square of the speed divided by the normal acceleration.

Answer: (a) \(m\omega^2a\)

Explanation:
1. A centrifugal force is a pseudo force that depends only on the angular velocity of the rotating frame and on the position of the particle in it.
2. The distance of the particle from the axis is a.
3. So the centrifugal force is \(m\omega^2a\), and it does not depend on \(\omega_0\).

Teacher's Note:
1. Centrifugal force belongs to the frame of reference, not to the motion of the particle.
2. It is always directed away from the axis of rotation.

Answer: (b) quadrupled

Explanation:
1. The tendency to overturn is measured by the centrifugal force \(\frac{mv^2}{r}\).
2. This is proportional to \(v^2\).
3. Doubling the speed makes it four times as large.

Teacher's Note:
1. The centripetal (or centrifugal) force varies as the square of the speed.
2. Slow down before a sharp turn.

Answer: (a) the top of the circle

Explanation:
1. At the top, \(T+mg=\frac{mv^2}{r}\), so \(T=16-10=6\) N.
2. At the bottom, \(T-mg=\frac{mv^2}{r}\), so \(T=26\) N, which does not match.
3. So the tension is 6 N at the top of the circle.

Teacher's Note:
1. At the top both the tension and the weight point towards the centre.
2. At the bottom the tension must overcome the weight as well.

Answer: (b) 3g

Explanation:
1. The critical speed at the top is \(v=\sqrt{rg}\).
2. Between the top and the horizontal position the mass falls through a height r: \(v'^2=v^2+2gr=3gr\).
3. The centripetal acceleration is \(\frac{v'^2}{r}=3g\).

Teacher's Note:
1. The critical speed at the top of a vertical circle is \(\sqrt{rg}\).
2. Apply energy conservation to find the speed at the new position.

Answer: (d) \(\frac{2as^2}{R}\)

Explanation:
1. \(K=\frac12mv^2=as^2\), so \(mv^2=2as^2\).
2. The centripetal force is \(\frac{mv^2}{R}\).
3. \(F=\frac{2as^2}{R}\).

Teacher's Note:
1. Rewrite the kinetic energy to get the combination mv².
2. The centripetal force is mv² divided by the radius.

Answer: (a) 0.75

Explanation:
1. Friction supplies the centripetal force, so \(\frac{mv^2}{r}\le\mu mg\).
2. \(\mu\ge\frac{v^2}{rg}\).
3. \(\mu_{min}=\frac{26^2}{92\times9.8}=\frac{676}{901.6}\approx0.75\).

Teacher's Note:
1. The smallest coefficient is found when friction is just at its limit.
2. The mass of the car cancels.

Answer: (a) \(\alpha=\tan^{-1}\left(\frac{2l}{r}\right)\)

Explanation:
1. The tangential acceleration is \(a_t=\frac{dv}{dt}=\frac{a}{2\sqrt l}\frac{dl}{dt}=\frac{av}{2\sqrt l}=\frac{a^2}{2}\).
2. The normal acceleration is \(a_n=\frac{v^2}{r}=\frac{a^2l}{r}\).
3. The velocity is along the tangent, so \(\tan\alpha=\frac{a_n}{a_t}=\frac{a^2l/r}{a^2/2}=\frac{2l}{r}\).

Teacher's Note:
1. The angle with the velocity is found from the ratio of the normal and tangential components.
2. Use \(\frac{dl}{dt}=v\) while differentiating.

Answer: (b) \(\sqrt{2gl}\)

Explanation:
1. To reach the level of the point of suspension the sphere rises through a height l.
2. At that point the thread is horizontal and can be slack with zero speed, so \(\frac12mu^2=mgl\).
3. \(u=\sqrt{2gl}\).

Teacher's Note:
1. Reaching the level of the support needs only the speed to rise a height l, not a full loop.
2. At the level of the support the thread is horizontal and the speed is zero, so the thread has just stopped being stretched.

Answer: (d) 96 rad

Explanation:
1. The angular acceleration is \(\alpha=\frac{24-0}{8}=3\ \text{rad/s}^2\).
2. \(\theta=\omega_0t+\frac12\alpha t^2=\frac12\times3\times8^2\).
3. \(\theta=96\) rad.

Teacher's Note:
1. Use the rotational equations of motion, which have the same form as the linear ones.
2. Check with the average angular velocity: \(12\times8=96\).

Answer: (a) track 1

Explanation:
1. The entry speed and the height of the top are the same for all tracks, so by energy conservation the speed at the top is the same in all of them.
2. At the highest point \(mg+N=\frac{mv^2}{R}\), so \(N=\frac{mv^2}{R}-mg\).
3. N is largest where the radius of curvature R is smallest, that is in track 1.

Teacher's Note:
1. A smaller radius of curvature needs a larger centripetal force at the same speed.
2. The speed at the top depends only on the height, not on the shape of the track.