for 6.1 The combination of an applied force and a friction force pr During this

Question

for 6.1 The combination of an applied force and a friction force pr During this time, the angular speed of the wheel increases from 0 to 10.1 rad/s. The applied force is then removed, and the wheel comes to rest in 60.5 s. (a) Find the moment of inertia of the wheel. kg m2 Find the magnitude of the torque due to friction. N M (c) Find the total number of revolutions of the wheel during the entire Interval of 66.6 s revolutions In the figure below, the hanging object has a mass of m - 0.465 kg; the sliding black has a mass of m2 0.840 kg: and the pulley Is a hollow cylinder with a mass of M - 0.350 kg, an inner radius of R 0.020 0 m, and an outer radius of R2 -o.030 between the block and the horizontal surface is -0.250. The pulley turns without friction on its axie. The light cord does not stretch and does not slip on th block has a velocity of v 0.820 m/s toward the pulley when it passes a reference point on the ta 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction 1n1 (a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away (b) Find the angular speed of the pulley at the same moment.

Explanation / Answer

Given,

Tau = 36.1 N-m ; t = 6.1 N ; alpha = 10.1 rad/s ; t' = 60.5 s

a)we know that

Tau = I alpha

alpha = w/t = 10.1/6.1 = 1.66 rad/s^2

I = Tau/alpha = 36.1/1.66 rad/s^2 = 21.75 kg-m^2

Hence, I = 21.75 kg-m^2

b)In this case

alpha = 10.1/60.5 = 0.167 rad/s^2

I = 36.1/0.167 = 216.25 kg-m^2

Hence, I = 216.25 kg-m^2

c)We know from eqn in circular motion

theta = wt + 1/2 alpha t^2

theta = 0.5 x 1.66 x 6.1^2 = 30.88

theta' = 10.1 x 60.5 - 0.5 x 0.167 x 60.5^2 = 305.42 rad

theta-f = 30.88 + 305.42 = 336.3

N = 336.3/2 pi = 53.55 rev

Hence, N= 53.55 rev

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