In the figure on the left, a pulley is driven by a \"timing belt\". The bumps on

Question

In the figure on the left, a pulley is driven by a "timing belt". The bumps on the belt prevent slipping between the belt and the pulley, and this is used in machines when shaft rotations must be coordinated. The effective outside diameter of the pulley is given by the "pitch diameter" denoted Phi dp=12cm. If the pulley starts from rest and is given an angular acceleration of alpha (t) = 1.23 - e-0.7t, determine the angular acceleration at t=5s, the angular velocity of the pulley at t=5s, and the linear distance of belt that has passed over the pulley since t=0.

Explanation / Answer

a(t) = 1.23 - e^(-0.7t)

at t= 5sec

a(t) = 1.23 - e^(-0.7 *5)

= 1.199 s^(-2)

let angular velocity be v

now dv/dt = a(t)

therefore v = integration(a(t) dt)

= 1.23*t + e^(-0.7t)/0.7

at t =5 sec

v= 1.23*5 +e^(-3.5)/0.7

=6.1931 s^(-1)

linear velocity V= v*radius

12*(1.23*t + e^(-0.7t)/0.7)

now s be the distance

ds/dt = V(t)

s= integration(V(t)*dt)

=12*((1.23*t^2)/2 - e^(-0.7t)/0.49)

put t=5sec

s= 183.76 cm

=1.8376 m

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