Question

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?

Answer 1

Hope it helps.

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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