Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 348: 4

(a) $a = 0.057~m/s^2$ (b) The length of the chain that passes over the sprocket is 7.85 meters.

Work Step by Step

(a) Initially, the crank arm rotates at 60 rpm which is 1 rev/s which is $2\pi~rad/s$. Therefore $\omega_0 = 2\pi~rad/s$ After ten seconds, the crank arm rotates at 90 rpm which is 1.5 rev/s which is $3\pi~rad/s$. Therefore $\omega_f = 3\pi~rad/s$ We can find the angular acceleration of the crank arm. $\alpha = \frac{\omega_f-\omega_0}{t}$ $\alpha = \frac{3\pi~rad/s-2\pi~rad/s}{10~s}$ $\alpha = 0.314~rad/s^2$ We can find the tangential acceleration of the pedal. $a = \alpha~r$ $a = (0.314~rad/s^2)(0.18~m)$ $a = 0.057~m/s^2$ (b) We can find $\theta$ during the 10-second acceleration period. Note that the angular acceleration of the sprocket is the same as the angular acceleration of the crank arm. $\theta = \omega_0~t+\frac{1}{2}\alpha~t^2$ $\theta = (2\pi~rad/s)(10~s)+\frac{1}{2}(0.314~rad/s^2)(10~s)^2$ $\theta = 78.5~rad$ We can find the distance $x$ that a point on the edge of the sprocket rotates. Note that this distance is equal to the length of the chain that passes over the sprocket. $x = \theta ~r$ $x = (78.5~rad)(0.10~m)$ $x = 7.85~m$ The length of the chain that passes over the sprocket is 7.85 meters.

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