A 1.7 kg solid disk pulley of radius 0.18 m rotates about an axis through its ce

Question

A 1.7 kg solid disk pulley of radius 0.18 m rotates about an axis through its center. (a) What is its moment of inertia? (b) Starting from rest, a torque 0.42 N-m will produce what angular acceleration? A mass of 0.410 kg is attached to a spring and undergoes simple harmonic oscillations with a period of 0.620 s. What is the force constant of the spring? A) 53.1 N/m B) 0.610 N/m c) 24.1 N/m D) 2.45 N/m E) 42.1 N/m A pendulum is observed to have a period of oscillation of 10.6 s. What was the length of the pendulum? A) 21 m B) 39 m C) 28 m D) 5 m E) 17 m

Explanation / Answer

8) m = 1.7 kg , r = 0.18 m

moment of inertia of disk I = 0.5mr^2

(a) I = 0.5*1.7*0.18^2

I = 0.0275 kg.m^2

(b) Torque = 0.42 N.m

Torque = I*alpha

0.42 = 0.0275*alpha

alpha = 15.3 rad/s^2

9) m = 0.41 kg , T = 0.62 s

T = 2pi(m/k)^0.5

0.62 = 6.28*(0.41/k)^0.5

k = 42.1 N/m

correct option is (E)

10) T = 10.6 s

T = 2pi(L/g)^0.5

10.6 = 6.28*(L/9.8)^0.5

L = 28 m

correct option is (C)

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