Question

Consider the 12 kg motorcycle wheel shown in the figure. Assume it to be approximately a ring with an inner radius of 0.255 m and an outer radius of 0.32 m. The motorcycle is on its center stand, so that the wheel can spin freely.

33% Part (a) If the drive chain exerts a force of 2250 N at a radius of 4.8 cm, what is the angular acceleration of the wheel, in radians per square second?       33% Part (b) What is the tangential acceleration, in meters per square second, of a point on the outer edge of the tire?   33% Part (c) How long, in seconds, starting from rest, does it take to reach an angular velocity of 83.5 rad/s?

Answer 1

Mass of the motorcycle wheel = M = 12 kg

Inner radius of the wheel = R₁ = 0.255 m

Outer radius of the wheel = R₂ = 0.32 m

Moment of inertia of the wheel = I

Moment of inertia of an annular disc is given by,

I = 1.00455 kg.m²

Force exerted on the wheel by the drive chain = F = 2250 N

Radius at which the drive chain exerts the force = R = 4.8 cm = 0.048 m

Angular acceleration of the wheel =

= 107.5 rad/s²

Tangential acceleration of a point on the outer edge of the tire = a_t

a_t = 34.4 m/s²

Initial angular velocity of the wheel = ₁ = 0 rad/s (At rest)

Final angular velocity of the wheel = ₂ = 83.5 rad/s

Time taken by the wheel to reach 83.5 rad/s = T

T = 0.777 sec

a) Angular acceleration of the wheel = 107.5 rad/s²

b) Tangential acceleration of a point on the outer edge of the tire = 34.4 m/s²

c) Time taken by the wheel to reach an angular velocity of 83.5 rad/s = 0.777 sec