Question 21:
easyA cyclist paddling at a speed of 10 m/s on a level road takes a sharp circular turn of radius 10 m without reducing the speed. The angle made by cyclist with vertical is
Using the formula
and given values
m/s,
m, and
m/s², we get:
Thus,
. The cyclist leans at 45° or π/4 with the vertical.
Question 22:
A particle is executing uniform circular motion with velocity \(\vec{v}\) and acceleration \(\vec{a}\). Which of the following is true?
In uniform circular motion, the magnitudes of velocity and centripetal acceleration are constant, but their directions continuously change as the particle moves along the circle. Thus, both vector quantities are non-constant.
Question 23:
easyA particle moves in a circular path so that its distance travel varies with time \(t\) as \(s = 3t^2 + 6t\). Then its acceleration at \(t = 1\text{ sec.}\) is (radius of path is \(12\text{ m}\)) –
Speed is \(v = \frac{ds}{dt} = 6t + 6\). At \(t = 1\text{ s}\), \(v = 12\text{ m/s}\). Tangential acceleration is \(a_t = \frac{dv}{dt} = 6\text{ m/s}^2\). Centripetal acceleration is \(a_c = \frac{v^2}{R} = \frac{12^2}{12} = 12\text{ m/s}^2\). Total acceleration is \(a = \sqrt{a_t^2 + a_c^2} = \sqrt{6^2 + 12^2} = 6\sqrt{5}\text{ m/s}^2\).
Question 24:
moderateThe kinetic energy \((K)\) of particle moving along a circle of radius \(R\) depends upon the distance covered \(S\) and is given by \(K = aS\) where \(a\) is a constant. Then the centripetal force acting on the particle is:
Kinetic energy is \(K = \frac{1}{2}mv^2 = aS\). Centripetal force is \(F_c = \frac{mv^2}{R}\) ( Since \(mv^2 = 2aS\), we get \(F_c = \frac{2aS}{R}\).
Question 25:
moderateA particle of mass \(m\) is tied to a string of length \(L\) and rotated in vertical circle about other end with critical speed so that it is just able to complete the vertical loop. Then tension in string, when string is at horizontal position will be:
To just complete the vertical loop, the velocity at the bottom is \(\sqrt{5gL}\). By energy conservation, the velocity at the horizontal position is \(v = \sqrt{3gL}\). The tension at this point is \(T = \frac{mv^2}{L} = 3mg\).
Question 26:
easyA particle start revolving on a circular path with constant angular acceleration \(\frac{\pi}{2}\text{ rad/sec}^2\). Then find number of cycles it will complete in first 12 seconds:
Angular displacement is \(\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2} \left(\frac{\pi}{2}\right) (12)^2 = 36\pi\text{ rad}\). Number of cycles \(N = \frac{\theta}{2\pi} = \frac{36\pi}{2\pi} = 18\).
Question 27:
easyAssertion: In uniform circular motion tangential acceleration of particle is zero.
Reason: In uniform circular motion net force on particle is always directed towards centre of circular path.
Tangential acceleration is zero because speed is constant. The net force is centripetal, which is directed towards the center. Thus, both are true but the Reason is not the explanation of the Assertion.
Question 28:
easyA road of width \(20\text{ m}\) forms an arc of radius \(15\text{ m}\), its outer edge is \(2\text{ m}\) higher than its inner edge. For what speed the road is banked?
The angle of banking is given by \(sin\theta \approx \tan\theta = \frac{h}{w} = \frac{2}{20} = 0.1\). Also, \(\tan\theta = \frac{v^2}{Rg}\). Equating these, \(frac{v^2}{15 \times 9.8} = 0.1 ⇒ v^2 = 14.7 ⇒ v = \sqrt{14.7}\text{ m/s}\).
Question 29:
easyA particle moves in a circle of radius \(10\text{ cm}\) with constant speed and time period \(\pi\text{ s}\). The acceleration of the particle is :
The angular frequency is \(\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2\text{ rad/s}\). The centripetal acceleration is \(a = \omega^2 R = 2^2 \times 10 = 40\text{ cm/s}^2\).
Question 30:
easyAssertion (A): A particle moving at constant speed and constant magnitude of radial acceleration must be undergoing uniform circular motion.
Reason (R): In uniform circular motion speed cannot change as there is no tangential acceleration.
Assertion (A): Constant speed and constant magnitude of radial acceleration ((v^2/r)) imply constant radius ((r)), which defines uniform circular motion. So (A) is True.
Reason (R): In uniform circular motion, acceleration is purely centripetal (radial), with no component tangential to the path. Thus, speed remains constant. So (R) is True.
Reason (R) correctly explains why constant speed and constant radial acceleration magnitude lead to uniform circular motion by implying constant radius and absence of tangential acceleration.