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IBDP Physics- A.4 Rigid body mechanics- IB Style Questions For HL Paper 1A -FA 2025

IBDP Physics HL Paper 1 -FA 2025- All Chapters

Question 

An ice skater is spinning with their arms extended in a fixed position at a constant angular velocity. The ice skater then quickly pulls their arms closer to their body. Frictional effects are negligible.

Three statements are made about the ice skater’s motion.

I. The angular momentum of the ice skater remains constant.

II. The rotational kinetic energy of the ice skater remains constant.

III. The net torque acting on the ice skater is zero.

Which of the statements are correct?

(A) I and II only
(B) I and III only
(C) II and III only
(D) I, II and III
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{B}} \)

Because frictional effects are negligible, there is no external torque acting on the ice skater about the rotation axis.

The relationship between torque and angular momentum is

\(\tau=\dfrac{\Delta L}{\Delta t}\)

Since the net external torque is zero, the angular momentum remains constant:

\(L=I\omega=\mathrm{constant}\)

Therefore, statement I is correct.

When the skater pulls their arms closer to their body, their moment of inertia \(I\) decreases. Since \(I\omega\) remains constant, the angular velocity \(\omega\) increases.

The rotational kinetic energy is

\(K_{\mathrm{rot}}=\dfrac{1}{2}I\omega^2\)

Since \(I\) decreases while \(\omega\) increases, the rotational kinetic energy does not remain constant. The skater does work by pulling their arms inward, increasing the rotational kinetic energy.

Therefore, statement II is incorrect.

The net external torque is zero because frictional effects are negligible. Therefore, statement III is correct.

Thus, the correct statements are I and III only.

Hence, the correct answer is \( \boxed{\mathrm{B}} \).

Question 

The graph shows how the angular acceleration \(\alpha\) of a flywheel varies with torque \(\tau\) applied to the flywheel.

What is the moment of inertia of the flywheel?

(A) \(0.20\,\mathrm{kg\,m^2}\)
(B) \(5.0\,\mathrm{kg\,m^2}\)
(C) \(40\,\mathrm{kg\,m^2}\)
(D) \(80\,\mathrm{kg\,m^2}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{A}} \)

For rotational motion, the relationship between torque, moment of inertia and angular acceleration is

\(\tau=I\alpha\)

Rearranging,

\(\alpha=\frac{\tau}{I}\)

Therefore, the gradient of the graph of \(\alpha\) against \(\tau\) is

\(\mathrm{gradient}=\frac{\alpha}{\tau}=\frac{1}{I}\)

From the graph, when \(\tau=4\,\mathrm{N\,m}\), \(\alpha=20\,\mathrm{rad\,s^{-2}}\).

Thus,

\(\frac{1}{I}=\frac{20}{4}=5\)

Hence,

\(I=\frac{1}{5}=0.20\,\mathrm{kg\,m^2}\)

Therefore, the moment of inertia of the flywheel is

\( \boxed{I=0.20\,\mathrm{kg\,m^2}} \)

Hence, the correct answer is \( \boxed{\mathrm{A}} \).

Question


The center of mass of a cylinder of mass \(m\), radius \(r\), and rotational inertia \(I=\tfrac{1}{2}mr^2\) has a velocity of \(v_{cm}\) as it rolls without slipping along a horizontal surface. It then encounters a ramp of angle \(\theta\), and continues to roll up the ramp without slipping.

What is the maximum height the cylinder reaches?

(A) \( \dfrac{v^2}{4g} \)
(B) \( \dfrac{4v^2}{3g} \)
(C) \( \dfrac{v^2}{3g} \)
(D) \( \dfrac{3v^2}{4g} \)
(E) \( \dfrac{4g}{3v^2} \)
▶️ Answer/Explanation
Detailed solution

Use conservation of mechanical energy.

Initial energy (rolling without slipping) is the sum of translational and rotational kinetic energies:
\(K_T=\tfrac{1}{2}mv_{cm}^2\), and \(K_R=\tfrac{1}{2}I\omega^2\).

For rolling without slipping, \(v_{cm}=\omega r\) so \(\omega=\dfrac{v_{cm}}{r}\).

At the maximum height, the cylinder momentarily comes to rest, so all kinetic energy becomes gravitational potential energy \(mgh\):
\(\tfrac{1}{2}mv_{cm}^2+\tfrac{1}{2}I\omega^2=mgh\).

Substitute \(I=\tfrac{1}{2}mr^2\) and \(\omega=\dfrac{v_{cm}}{r}\):
\(\tfrac{1}{2}mv_{cm}^2+\tfrac{1}{2}\left(\tfrac{1}{2}mr^2\right)\left(\dfrac{v_{cm}}{r}\right)^2=mgh\).
\(\tfrac{1}{2}mv_{cm}^2+\tfrac{1}{4}mv_{cm}^2=mgh\).
\(\tfrac{3}{4}mv_{cm}^2=mgh\).

Therefore \(h=\dfrac{3v_{cm}^2}{4g}\), which corresponds to option (D).

✅ Answer: (D)

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