Consider the 12 kg motorcycle wheel shown in the figure. Assume it to be approxi

Question

Consider the 12 kg motorcycle wheel shown in the figure. Assume it to be approximately a ring with an inner radius of 0.27 m and an outer radius of 0.35 m. The motorcycle is on its center stand, so that the wheel can spin freely. Part (a) If the drive chain exerts a force of 2050 N at a radius of 5. 1 cm, what is the angular acceleration of the wheel, in radians per square second? Part (b) What is the tangential acceleration, in meters per square second, of a point on the outer edge of the tire?

Explanation / Answer

tangential accelaration is a_tan = R*alpha

R = 0.35 m

alpha is the angular accelaration

Torque due to Force is T = r*F

but T = I*alpha = r*F

I is the moment of inertia 0.5*m*(r^2+R^2) = 0.5*12*(0.27^2+0.35^2) = 1.1724 kg-m^2

alpha is the angular accelaration

r = 5.1 cm = 0.051 m

F = 2050 N

then

I*alpha = r*F

1.1724*alpha = 0.051*2050

alpha = 89.17 rad/s^2

then a_tan = R*alpha= 0.35*89.17 = 31.21 m/s^2

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