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/sec^2)

(b) What is the tangential acceleration of a point on the outer edge of the tire? (m/sec^2)

(c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s? (rad/sec)

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/sec^2) (b) What is the tangential acceleration of a point on the outer edge of the tire? (m/sec^2) (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s? (rad/sec)

Explanation / Answer

momet of Inertia of the wheel, I = 0.5*M*(R1^2 + R2^2)

= 0.5*12*(0.28^2+0.36^2)

= 1.248 kg.m^2

Torque, T = F*r*sin(90)

= 2100*0.05*1

= 105 N.m

a) Apply, T = I*alfa

==> alfa = T/I

= 105/1.248

= 83.13 rad/s^2

b) a_tan = R2*alfa

= 0.36*84.13

= 30.3 m/s^2

c) Apply, w = wo + alfa*t

==> t = (w-wo)/alfa

= (80 - 0)/83.13

= 0.962 s

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