Life is like riding a bicycle.  To keep your balance you must keep moving.”
Albert Einstein

Thursday, January 24, 2008

AP Physics B and C– Multiple Choice Questions on Circular Motion and Rotation

The following questions will be of common interest to AP Physics B as well as AP Physics C aspirants:

(1) A fighter pilot flies along a vertical circle of radius 1 km. At the topmost point of the circle, he feels weightless. What is the speed of the fighter plane? (Acceleration due to gravity = 10 ms–2)

(a) 25 ms–1

(b) 50 ms–1

(c) 75 ms–1

(d) 100 ms–1

(e) 125 ms–1

He feels weightless since his weight is balanced by the centrifugal force so that we have

mv2/R = mg where m is the mass of the plane (with the pilot), R is the radius of the circle and v is the speed of the plane at the topmost point.

Therefore, v = √(Rg) = √(1000×10) = 100 ms–1.

(2) A door can be opened by pushing with a force of at least 50 N at a point distant 80 cm from the line of the hinges. The minimum force required to open the door by pushing at another point distant 20 cm from the line of the hinges is

(a) 200 N

(b) 100 N

(c) 50 N

(d) 25 N

(e) 12.5 N

The minimum torque (moment of force) required to open the door is given by

τ = r × F

The magnitude of the torque is τ = rF sin 90º, since the force is to be applied perpendicular to the plane of the door to produce maximum torque (so that the applied force is minimum).

Since the minimum torque required is the same in both cases, we have

0.8×50 = 0.2×F where F is the force required to open the door by pushing at a point distant 20 cm from the line of the hinges.

This gives F = 200 N.

(3) A wheel of radius R is rolling along a horizontal surface with a speed u. A pebble trapped on the wheel gets separated from the highest point of the wheel when the wheel arrives at position P (fig.). The horizontal range PQ of the pebble is

(a) u√(R/g)

(b) 2u√(R/g)

(c) 4u√(R/g)

(d) u2√(R/g)

(e) 2u2√(R/g)

Considering the vertical motion of the projectile (pebble), the time taken to reach the ground after leaving the wheel is given by

2R = 0 + (½)gt2, from which t = 2√(R/g)

[We have used the equation of one dimensional motion, x = x0 + v0t + (½) at2]

The horizontal range of the pebble is PQ = horizontal velocity×time of flight

The centre of mass of the wheel is moving with speed u. The pebble at the topmost point of the wheel is moving horizontally with speed u with respect to the wheel so that the horizontal velocity of the pebble with respect to the ground is u+u = 2u.

Therefore, horizontal range PQ = 2u ×2√(R/g) = 4u√(R/g)

In the next post we will discuss questions in this section solely for the benefit of AP Physics C aspirants.

Example isn't another way to teach, it is the only way to teach.

Albert Einstein

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