Consider the 12.0-kg motorcycle wheel shown in the figure below. Assume it to be

Question

Consider the 12.0-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 2100 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?
s

Consider the 12.0-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 2100 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? s

Explanation / Answer

Angular Velocity (Omega) = Angular displacement / time. You know that 1 revolution = 2 Pi radians. You have 2 revolutions = 4 x Pi radians in 1 second. Omega = 4Pi r/s OR 12.566 r/s

Tangential Velocity (v) = Omega x Radius. R = 0.3m. Thus v = 3.77m/s

For a MASS rotating on a chord of radius 0.3m at Omega 12.566r/s the Centrifugal Force = Mass x Omega^2 x Radius [ Force = Mass x Acceleration ]
Hence the acceleration Centrifugal = [ Omega^2 x Radius ] = 47.37 m/s^2
Centrifugal ACTS away from the centre of rotation WHEREAS Centripetal ACTS towards (OPPOSITE)
Hence the acceleration Centripetal = - 47.37 m/s^2

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