A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless

Question

A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest at the horizontal position. What is the angular acceleration of the rod at the instant the rod makes an angle of 70 degree with the horizontal? a. 3.7 rad/s^2 b. 1.3 rad/s^2 c. 2.5 rad/s^2 d. 4.9 rad/s^2 e. 1.9 rad/s^2 Two cylinders made of the same material roll down a plane inclined at an angle theta with the horizontal. Each travels the same distance. The radius of cylinder B is twice the radius of cylinder A. In what order do they reach the bottom? a. A reaches the bottom first because it has the greater acceleration. b. A reaches the bottom first because it has a smaller moment of inertia. c. B reaches the bottom first because is experiences a larger torque. d. B reaches the bottom first because it travels a larger distance in one rotation. e. They both reach the bottom at the same time, because each has the same linear acceleration. The angular speed of the hour hand of a clock, in rad/min, is a. 1/1800 pi. b. 1/60 pi. C. 1/30 pi d. pi e. 120 pi.

Explanation / Answer

4. Applying energy conservation,

PEi + KEi = PEf + KEf

0 + 0 = - m g (L/2) sin70 + I w^2 /2

m g L sin70 / 2 = (m L^2 / 3) w^2 / 2

3 x 9.8 x sin70 / 2 = w^2

w = 3.7 rad/s

Ans(a)

5. along the incline,

m g sin(theta) - f = m a

and f R = I (a / R )

I = m R^2 / 2 = (pi R^2 L ) R^2 /2 = 0.5 pi R^3 L

f R = (0.50 pi R^3 L ) (a / R)

f = 0.50 pi R^2 L a

putting in prvious equation,

(pi R^2 L ) g sin(theta) - 0.50 pi R^2 L a = (pi R^2 L ) a

(pi R^2 L ) g sin(theta) = 1.50 pi R^2 L a

a = g sin(theta) / 1.50

so a is independent of R hence both will reach at same time

Ans(e)

6. Time period of hour hand = 12 hr

w = 2pi rad / 12 hr = pi / 6 rad/hr

w = (pi / 6) rad / (60 min) = pi / 360 rad/min

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