Rolling Motion — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Rolling Motion MCQs with step-by-step solutions (21 questions). Part of Rotational Dynamics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Rolling Motion · easy · theory
The condition for a wheel of radius $r$ to roll without slipping is:
A. $v = \omega r$ ✓ Correct
B. $v^2 = \omega r$
C. $v = \dfrac{\omega}{r}$
D. $v = \omega^2 r$
Solution: Rolling without slipping means the contact point is instantaneously at rest, which requires the centre-of-mass speed to satisfy $v = \omega r$.
Q2 — Rolling Motion · medium · numerical
A solid sphere rolls without slipping. The ratio of its rotational kinetic energy to its translational kinetic energy is:
A. $1 : 2$
B. $5 : 2$
C. $1 : 1$
D. $2 : 5$ ✓ Correct
Solution: $\dfrac{K_{rot}}{K_{trans}} = \dfrac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2} = \dfrac{k^2}{R^2} = \dfrac{2}{5}$ for a solid sphere.
Q3 — Rolling Motion · medium · numerical
A solid sphere rolls without slipping down an incline. The fraction of its total kinetic energy that is rotational is:
A. $\dfrac{2}{7}$ ✓ Correct
B. $\dfrac{1}{2}$
C. $\dfrac{2}{5}$
D. $\dfrac{5}{7}$
Solution: The rotational fraction is $\dfrac{k^2/R^2}{1 + k^2/R^2} = \dfrac{2/5}{1 + 2/5} = \dfrac{2/5}{7/5} = \dfrac{2}{7}$.
Q4 — Rolling Motion · medium · numerical
A thin ring rolls without slipping on a horizontal surface. The ratio of its rotational kinetic energy to its total kinetic energy is:
A. $1 : 1$
B. $1 : 2$ ✓ Correct
C. $1 : 3$
D. $2 : 3$
Solution: For a ring $I = mR^2$, so $K_{rot} = \dfrac{1}{2}mv^2 = K_{trans}$ and the total is $mv^2$. Hence $K_{rot} : K_{total} = 1 : 2$.
Q5 — Rolling Motion · medium · numerical
A uniform disc rolls without slipping. The fraction of its total kinetic energy that is rotational is:
A. $\dfrac{1}{2}$
B. $\dfrac{2}{5}$
C. $\dfrac{1}{3}$ ✓ Correct
D. $\dfrac{2}{3}$
Solution: For a disc $\dfrac{k^2}{R^2} = \dfrac{1}{2}$, so the rotational fraction is $\dfrac{1/2}{1 + 1/2} = \dfrac{1}{3}$.
Q6 — Rolling Motion · easy · numerical
A sphere of radius $0.25\text{ m}$ rolls without slipping with a centre-of-mass speed of $5\text{ m/s}$. Its angular velocity is:
A. $20\text{ rad/s}$ ✓ Correct
B. $1.25\text{ rad/s}$
C. $5\text{ rad/s}$
D. $12.5\text{ rad/s}$
Solution: $\omega = \dfrac{v}{r} = \dfrac{5}{0.25} = 20\text{ rad/s}$.
Q7 — Rolling Motion · medium · theory
A solid cylinder rolls without slipping down an incline of angle $\theta$. Its linear acceleration is:
A. $\dfrac{1}{2}g\sin\theta$
B. $\dfrac{2}{3}g\sin\theta$ ✓ Correct
C. $g\sin\theta$
D. $\dfrac{5}{7}g\sin\theta$
Solution: $a = \dfrac{g\sin\theta}{1 + k^2/R^2}$. For a solid cylinder $\dfrac{k^2}{R^2} = \dfrac{1}{2}$, giving $a = \dfrac{g\sin\theta}{3/2} = \dfrac{2}{3}g\sin\theta$.
Q8 — Rolling Motion · medium · theory
A solid sphere rolls without slipping down an incline of angle $\theta$. Its linear acceleration is:
A. $\dfrac{3}{5}g\sin\theta$
B. $\dfrac{5}{7}g\sin\theta$ ✓ Correct
C. $\dfrac{2}{3}g\sin\theta$
D. $g\sin\theta$
Solution: With $\dfrac{k^2}{R^2} = \dfrac{2}{5}$, $a = \dfrac{g\sin\theta}{1 + 2/5} = \dfrac{g\sin\theta}{7/5} = \dfrac{5}{7}g\sin\theta$.
Q9 — Rolling Motion · hard · numerical
A solid sphere starts from rest and rolls without slipping down a height $h$. Its speed at the bottom is:
A. $\sqrt{\dfrac{4gh}{3}}$
B. $\sqrt{gh}$
C. $\sqrt{2gh}$
D. $\sqrt{\dfrac{10gh}{7}}$ ✓ Correct
Solution: Energy conservation: $mgh = \dfrac{1}{2}mv^2\left(1 + \dfrac{k^2}{R^2}\right) = \dfrac{1}{2}mv^2\left(\dfrac{7}{5}\right)$, giving $v = \sqrt{\dfrac{10gh}{7}}$.
Q10 — Rolling Motion · hard · numerical
A solid cylinder rolls without slipping from rest down a height $h$. Its speed at the bottom is:
A. $\sqrt{gh}$
B. $\sqrt{\dfrac{10gh}{7}}$
C. $\sqrt{2gh}$
D. $\sqrt{\dfrac{4gh}{3}}$ ✓ Correct
Solution: $mgh = \dfrac{1}{2}mv^2\left(1 + \dfrac{1}{2}\right) = \dfrac{3}{4}mv^2$, so $v = \sqrt{\dfrac{4gh}{3}}$.
Q11 — Rolling Motion · medium · theory
A solid sphere, a solid cylinder and a ring of equal mass and radius are released together from the top of the same incline. The body that reaches the bottom first is:
A. The solid cylinder
B. The solid sphere ✓ Correct
C. The ring
D. All three arrive together
Solution: Acceleration $a = \dfrac{g\sin\theta}{1 + k^2/R^2}$ is largest for the smallest $\dfrac{k^2}{R^2}$. The values are $\dfrac{2}{5}$ (sphere), $\dfrac{1}{2}$ (cylinder) and $1$ (ring), so the sphere wins.
Q12 — Rolling Motion · medium · numerical
A disc of mass $2\text{ kg}$ rolls without slipping with a speed of $4\text{ m/s}$. Its total kinetic energy is:
A. $24\text{ J}$ ✓ Correct
B. $32\text{ J}$
C. $16\text{ J}$
D. $8\text{ J}$
Solution: Total $K = \dfrac{1}{2}mv^2\left(1 + \dfrac{k^2}{R^2}\right) = \dfrac{1}{2}(2)(16)\left(\dfrac{3}{2}\right) = 16 \times 1.5 = 24\text{ J}$.
Q13 — Rolling Motion · medium · theory
A ring of mass $m$ rolls without slipping with speed $v$. Its total kinetic energy is:
A. $\dfrac{1}{2}mv^2$
B. $mv^2$ ✓ Correct
C. $\dfrac{7}{10}mv^2$
D. $\dfrac{3}{4}mv^2$
Solution: For a ring $\dfrac{k^2}{R^2} = 1$, so $K = \dfrac{1}{2}mv^2(1 + 1) = mv^2$ — translational and rotational parts are equal.
Q14 — Rolling Motion · medium · theory
The acceleration of a body rolling without slipping down a given incline is independent of:
A. Its mass and radius ✓ Correct
B. The acceleration due to gravity
C. The angle of the incline
D. Its radius of gyration
Solution: In $a = \dfrac{g\sin\theta}{1 + k^2/R^2}$ the mass cancels entirely and only the ratio $\dfrac{k^2}{R^2}$ (a pure number set by shape) appears — not the radius itself.
Q15 — Rolling Motion · hard · theory
During rolling without slipping on a fixed surface, the work done by the force of static friction is:
A. Equal to the rotational kinetic energy
B. Positive and equal to the loss in potential energy
C. Zero ✓ Correct
D. Negative
Solution: The point of application of static friction is the contact point, which is instantaneously at rest. Since the displacement of that point is zero, friction does no work — which is why mechanical energy is conserved in ideal rolling.
Q16 — Rolling Motion · hard · theory
A body of radius of gyration $k$ and radius $R$ rolls without slipping down an incline of angle $\theta$. Its linear acceleration is:
A. $\dfrac{g\sin\theta}{1 - \dfrac{k^2}{R^2}}$
B. $g\sin\theta\left(1 + \dfrac{k^2}{R^2}\right)$
C. $\dfrac{g\sin\theta}{1 + \dfrac{k^2}{R^2}}$ ✓ Correct
D. $\dfrac{g\cos\theta}{1 + \dfrac{k^2}{R^2}}$
Solution: Combining $mg\sin\theta - f = ma$ with $fR = I\alpha$ and $a = \alpha R$, where $I = mk^2$, gives $a = \dfrac{g\sin\theta}{1 + k^2/R^2}$.
Q17 — Rolling Motion · medium · numerical
A ring of mass $4\text{ kg}$ rolls without slipping at $3\text{ m/s}$. Its total kinetic energy is:
A. $12\text{ J}$
B. $18\text{ J}$
C. $36\text{ J}$ ✓ Correct
D. $27\text{ J}$
Solution: For a ring $\dfrac{k^2}{R^2} = 1$, so $K = \dfrac{1}{2}mv^2(1 + 1) = mv^2 = 4 \times 9 = 36\text{ J}$.
Q18 — Rolling Motion · easy · numerical
A sphere of radius $0.35\text{ m}$ rolls without slipping with a centre-of-mass speed of $7\text{ m/s}$. Its angular velocity is:
A. $2.45\text{ rad/s}$
B. $0.05\text{ rad/s}$
C. $7\text{ rad/s}$
D. $20\text{ rad/s}$ ✓ Correct
Solution: $\omega = \dfrac{v}{r} = \dfrac{7}{0.35} = 20\text{ rad/s}$.
Q19 — Rolling Motion · hard · numerical
A solid sphere rolls without slipping down an incline of $30^\circ$. Its linear acceleration is ($g = 10\text{ m/s}^2$):
A. $3.57\text{ m/s}^2$ ✓ Correct
B. $3.33\text{ m/s}^2$
C. $5.00\text{ m/s}^2$
D. $7.14\text{ m/s}^2$
Solution: $a = \dfrac{g\sin\theta}{1 + k^2/R^2} = \dfrac{10 \times 0.5}{1 + 2/5} = \dfrac{5}{7/5} = \dfrac{25}{7} \approx 3.57\text{ m/s}^2$.
Q20 — Rolling Motion · medium · numerical
A disc of mass $3\text{ kg}$ rolls without slipping at $2\text{ m/s}$. Its total kinetic energy is:
A. $3\text{ J}$
B. $6\text{ J}$
C. $12\text{ J}$
D. $9\text{ J}$ ✓ Correct
Solution: $K = \dfrac{1}{2}mv^2\left(1 + \dfrac{1}{2}\right) = \dfrac{3}{4}mv^2 = \dfrac{3}{4}(3)(4) = 9\text{ J}$.
Q21 — Rolling Motion · hard · numerical
A hollow sphere rolls without slipping down an incline of $30^\circ$. Its linear acceleration is ($g = 10\text{ m/s}^2$):
A. $5.00\text{ m/s}^2$
B. $2.50\text{ m/s}^2$
C. $3.00\text{ m/s}^2$ ✓ Correct
D. $3.57\text{ m/s}^2$
Solution: For a hollow sphere $\dfrac{k^2}{R^2} = \dfrac{2}{3}$, so $a = \dfrac{g\sin\theta}{1 + 2/3} = \dfrac{5}{5/3} = 3\text{ m/s}^2$.