Question
Consider the 15.0 kg motorcycle wheel shown in the figure. Assume it to be approximately an...
Consider the 15.0 kg motorcycle wheel shown in the figure.
Assume it to be approximately an annular ring with an inner radius of
R1 = 0.280 m
and an outer radius of
R2 = 0.300 m.
The motorcycle is on its center stand, so that the wheel can spin freely.
(a)If the drive chain exerts a force of 1800 N at a radius of 5.00 cm, what is the angular acceleration (in rad/s2) of the wheel?
(b)What is the tangential acceleration (in m/s2) of a point on the outer edge of the tire?
(c)How long (in s), starting from rest, does it take to reach an angular velocity of 80.0 rad/s?
Answers
Hope it helps.
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Moment of inentia of a is given by I annular disc (here, wheel of bike) RM _R," I'm (R2 + R, 2) R₂² - R2 Im 2 - ) 0.05 X 1800 =90 Nm (a) I = MXF Now, I = IK > a = 2 - & * 15% (0.300? + O- + 0.2803) I= 1 x (0.09+0.0784) 2 = 1.263 - 71.26 rad /s² Hence, angular accelerations 90 1 283 6) Tangential oce, a 2 = 0.300x71.26 - 21.38 m/s2 (c) we know, wf = Witat > t = علیا - ولا & t = 1.123 S CS 80_o Ścanned with CamScanner 71.26
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