AP Physics 1: 7.2 Torque and Angular Acceleration – Exam Style questions with Answer- FRQ

 Question

A box of uniform density weighing 100 newtons moves in a straight line with constant speed along a horizontal surface. The coefficient of sliding friction is 0.4 and a rope exerts a force F in the direction of motion as shown above.
a. On the diagram below, draw and identify all the forces on the box.

b. Calculate the force F exerted by the rope that keeps the box moving with constant speed.

c. A horizontal force F’, applied at a height 5/3 meters above the surface as shown in the diagram above, is just sufficient to cause the box to begin to tip forward about an axis through point P. The box is 1 meter wide and 2 meters high. Calculate the force F’.

▶️Answer/Explanation

Ans:

a) FBD. Fn pointing up, Fg pointing down, fk applied to base of box pointing left
b) Constant speed → a=0.
Fnet = 0                 F – fk = 0               F – uFn = 0             F – (0.4)(100)=0                F = 40 N
c) The force F’ occurs at the limit point of tipping which is when the torque trying to tip it (caused by F’) is equal to the torque trying to stop it from tipping (from the weight) using the tipping pivot point of the bottom right corner of the box.

 

(F’)(5/3 m) = Fg (0.5m)
(F’) (5/3) = (100)(0.5)
F’ = 30 N

Question

The diagram below shows a beam of length 20.0 m and mass 40.0 kg resting on two supports placed at 5.0 m from each end.

A person of mass 50.0 kg stands on the beam between the supports. The reaction forces at the supports are shown.
(a) State the value of N1 + N2
(b) The person now moves toward the X end of the beam to the position where the beam just begins to tip and reaction force N1 becomes zero as the beam starts to leave the left support. Determine the distance of the girl from the end X when the beam is about to tip.

▶️Answer/Explanation

Ans:

(a) Simple application of Fnet(y) = 0                                                              N1 + N2 – mbg – mpg = 0
                                                                                                                                 N1 + N2 = (40)(9.8) + (50)(9.8) = 882 N

(b) 

Apply rotational equilibrium
(mbg) • r1 = (mpg) • r2
(40) (5m) = (50) (r)

                                                                  r = 4m from hinge

                                                                → 1 m from point X

 Question

A ball of mass m is attached by two strings to a vertical rod. as shown above. The entire system rotates at constant angular velocity ω about the axis of the rod.
a. Assuming ω is large enough to keep both strings taut, find the force each string exerts on the ball in terms of ω, m, g, R, and θ.
b. Find the minimum angular velocity, ωmin for which the lower string barely remains taut.

▶️Answer/Explanation

Ans:

a. Define a coordinate system with the x-axis directed to the vertical rod and the y-axis directed upwards and perpendicular to the first. Let T1 be the tension in the horizontal string. Let T2 be the tension in the string tilted upwards.
Applying Newton’s Second Law: \(\sum F_{x}\) = T1 + T2 sin θ = mω2R; \(\sum F_{y}\) = T2 cos θ – mg = 0
Solving yields: T2 = mg/cos θ and T1 = m(ω2R – g tan θ)
b. Let T1 = 0 and solving for ω gives ω = (g tanθ/R)1/2

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