A thin hoop with a mass of 2 kg and radius 0.4 m is rotating about an axis that

Question

A thin hoop with a mass of 2 kg and radius 0.4 m is rotating about an axis that is perpendicular to the plane of the hoop and through its center. Initially, the angular velocity of the hoop is +20 rad/s. At t = 0 s, a net force F is applied tangentially to the rim of the hoop. Two seconds later, the angular velocity of the hoop is +12 rad/s. a. What is the angular acceleration of the hoop? Indicate the direction with the sign of your answer. -4 rad/s^2 b. What is the net torque acting on the hoop. Indicate the direction with the sign of your answer. -1.28 N middot m c. What is the magnitude of the force F? 3.2 N d. What is the work done on the hoop between t = 0 and t = 2 s? -40.96 J

Explanation / Answer

Given,

Mass, M = 2 kg

Radius, r = 0.4 m

Initial angular speed, 1 = 20 rad/s

Final angular speed, 2 = 12 rad/s

a) Using Newton's first equation of motion,

f - i = t

So, angular acceleration, = (12-20) / 2 = - 4 rad/s2

b) Net torque acting on the hoop, = I = Mr2 x = 2 x 0.4 x 0.4 x (-4) = - 1.28 Nm

c) Magnitude of force, F = / r = 1.28 / 0.4 = 3.2 N

d) Using Newton's third equation of motion,

  f2 - i2  = 2

   = (122 - 202) / (2 x (-4)) = 32 rad

So, work done, W = x = 32 x (-1.28) = - 40.96 J

(Please rate my answer if you find it helpful, good luck...)

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