The drawing shows the top view of two doors. The doors are uniform and identical

Question

The drawing shows the top view of two doors. The doors are uniform and identical. Door A rotates about an axis through its left edge, and door B rotates about an axis through its center. The same force is applied perpendicular to each door at its right edge, and the force remains perpendicular as the door turns. No other force affects the rotation of either door. Starting from rest, door A rotates through a certain angle in 2.65 s. How long does it take door B (also starting from rest) to rotate through the same angle?

Explanation / Answer

Let L is the length and M is the mass of door.

In case A,

Moment of Inertia of the door, IA = M*L^2/3

Torque, TA = F*L

Let alfa_A is angular acceleration.

Apply, TA = IA*alfa_A

alfa_A = TA/IA

= F*L/(M*L^2/3)

= 3*F/(M*L) ---(1)

In case B,

Moment of Inertia of the door, IB = M*L^2/12

Torque, TB = F*(L/2)

Let alfa_B is angular acceleration.

Apply, TB = IA*alfa_B

alfa_B = TB/IB

= (F*L/2)/(M*L^2/12)

= 6*F/(M*L) ---(2)

Let tA and tB are the taken for door A and door B.

given angular dispalce ment is same

theta_B = theta_A

0.5*alfa_B*tB^2 = 0.5*alfa_A*tA^2

tB = tA*sqrt(alfa_A/alfa_B)

= 2.65*sqrt(6/3)

= 3.75 s <<<<<<<-----------Answer

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