Consider the 14.5-kg motorcycle wheel shown in the figure below. Assume it to be

Question

Consider the 14.5-kg motorcycle wheel shown in the figure below. Assume it to be approximately an annular ring with an inner radius of R1 = 0.280 m and an outer radius of R2 = 0.360 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If the drive chain exerts a force of 2275 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? rad/s2 (b) What is the tangential acceleration of a point on the outer edge of the tire? m/s2 (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s?

Explanation / Answer

a)

for an annular ring ,

moment of inertia , I = m*(R1^2 + R2^2 )/2

I = 14.5 *(0.28^2 + 0.36^2)/2

I = 1.508 Kg.m/s

Now , Using second law of motion ,

F*r= I*a

2275 * 0.05 = 1.508 * a

a = 75.43rad/s^2

the angular acceleration is 75.43rad/s^2

b)

at = a*R2

at = 75.43 * 0.36

at = 27.15 m/s^2

the tangential acceleration is 27.15 m/s^2

c)

Using first equation of motion

80 = 75.43 * t

t = 1.06 s

the tima taken is 1.06 s  

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