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AP Physics 1- 5.6 Newton’s Second Law in Rotational Form - Exam Style questions - FRQs- New Syllabus

Newton’s Second Law in Rotational Form AP  Physics 1 FRQ

Unit 5: Torque and Rotational Dynamics

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Question

The left end of a uniform beam of mass \( M \) and length \( L \) is attached to a wall by a hinge, as shown in Figure \( 1 \). One end of a string with negligible mass is attached to the right end of the beam, and the other end of the string is attached to the wall above the hinge at Point \( 1 \). The beam remains horizontal. The hinge exerts a force on the beam of magnitude \( F_H \), and the angle between the beam and the string is \( \theta = \theta_1 \).
(a) The following rectangle represents the beam in Figure \( 1 \). On the rectangle, draw and label the forces \( (\text{not components}) \) exerted on the beam. Draw each force as a distinct arrow starting on, and pointing away from, the point at which the force is exerted.
(b) The string is then attached lower on the wall, at Point \( 2 \), and the beam remains horizontal, as shown in Figure \( 2 \). The angle between the beam and the string is \( \theta = \theta_2 \). The dashed line represents the string shown in Figure \( 1 \).
 
The magnitude of the tension in the string shown in Figure \( 1 \) is \( F_{T1} \). The magnitude of the tension in the string shown in Figure \( 2 \) is \( F_{T2} \). Indicate which of the following correctly compares \( F_{T2} \) with \( F_{T1} \).
_____ \( F_{T2} > F_{T1} \)      _____ \( F_{T2} < F_{T1} \)      _____ \( F_{T2} = F_{T1} \)
Briefly justify your answer, using qualitative reasoning beyond referencing equations.
(c) Starting with Newton’s second law in rotational form, derive an expression for the magnitude of the tension in the string. Express your answer in terms of \( M \), \( \theta \), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference book.
(d) Is your derived equation in part \( \mathrm{(c)} \) consistent with your justification in part \( \mathrm{(b)} \)? Explain your reasoning.
(e) The string is cut, and the beam begins to rotate about the hinge with negligible friction. On the following axes, sketch the angular speed of the beam as a function of time for the time interval while the beam falls but before the beam becomes vertical.
 

Most-appropriate topic codes (AP Physics 1):

• Topic \( 2.2 \) — Forces and Free-Body Diagrams (Part \( \mathrm{(a)} \))
• Topic \( 5.4 \) — Torque and the Second Condition of Equilibrium (Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \), Part \( \mathrm{(d)} \))
• Topic \( 5.5 \) — Rotational Equilibrium and Newton’s First Law in Rotational Form (Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \))
• Topic \( 5.6 \) — Newton’s Second Law in Rotational Form (Part \( \mathrm{(c)} \), Part \( \mathrm{(e)} \))
• Topic \( 6.2 \) — Torque and Work (Part \( \mathrm{(e)} \))
▶️ Answer/Explanation

(a)
Three forces act on the beam:

\( \bullet \) The gravitational force \( F_g = Mg \), acting downward at the center of the beam
\( \bullet \) The tension force \( F_T \), acting at the right end of the beam along the string, directed upward and leftward
\( \bullet \) The hinge force \( F_H \), acting at the hinge

A correct free-body diagram therefore has one downward arrow at the center, one up-left arrow at the right end, and one force arrow at the hinge.

(b)
\( \boxed{F_{T2} > F_{T1}} \)

The beam remains horizontal and in rotational equilibrium in both cases, so the torque produced by the string must balance the torque produced by the beam’s weight.

When the string is attached lower on the wall, the angle \( \theta \) becomes smaller. That means the perpendicular component of the tension is smaller for the same tension. To produce the same balancing torque, the tension must therefore be larger.

So a smaller string angle requires a larger tension.

(c)
Start with Newton’s second law for rotation:

\( \sum \tau = I\alpha \)

Since the beam is in equilibrium,

\( \alpha = 0 \qquad \Rightarrow \qquad \sum \tau = 0 \)

Take torques about the hinge so the hinge force produces no torque.

The torque due to the tension has magnitude

\( \tau_T = (F_T \sin\theta)L \)

because the tension acts at the end of the beam, a distance \( L \) from the hinge, and only the perpendicular component contributes.

The torque due to the beam’s weight has magnitude

\( \tau_g = Mg\left(\dfrac{L}{2}\right) \)

because the weight acts at the center of mass of the uniform beam, halfway along its length.

Setting the net torque equal to zero:

\( (F_T \sin\theta)L – Mg\left(\dfrac{L}{2}\right) = 0 \)

\( (F_T \sin\theta)L = Mg\left(\dfrac{L}{2}\right) \)

Cancel \( L \):

\( F_T \sin\theta = \dfrac{Mg}{2} \)

Therefore,

\( \boxed{F_T = \dfrac{Mg}{2\sin\theta}} \)

This result shows that the tension depends inversely on \( \sin\theta \).

(d)
Yes, the derived equation is consistent with the reasoning in part \( \mathrm{(b)} \).

From \( F_T = \dfrac{Mg}{2\sin\theta} \), the tension is inversely proportional to \( \sin\theta \). When the string is attached lower, the angle decreases from \( \theta_1 \) to \( \theta_2 \), so \( \sin\theta \) becomes smaller.

A smaller denominator makes \( F_T \) larger, so the equation predicts \( F_{T2} > F_{T1} \), exactly matching the qualitative argument in part \( \mathrm{(b)} \).

(e)
The angular speed starts at \( 0 \) when the string is cut, then increases as the beam falls.

The graph should be monotonically increasing and concave down.

Early in the motion, the beam speeds up quickly because the gravitational torque is relatively large. As the beam approaches vertical, the lever arm of the weight decreases, so the torque and angular acceleration decrease. Therefore the angular speed continues to increase, but at a decreasing rate.

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