Class t | Help HW14 (Rotational motion) Begin Date: 10/31/2017 11:00:00 P M Due
ID: 1782121 • Letter: C
Question
Class t | Help HW14 (Rotational motion) Begin Date: 10/31/2017 11:00:00 P M Due Date: 11/14/2017 11:59.00 PM End Date: 1214/2017 11.59:00 PM (7%) Problem 11: Consider the 12 kg motorcycle wheel shown in the figure. Assume it to be approximately a ring with an inner radius of 0.29 m and an outer radius of 0.31 m. The motorcycle is on its center stand, so that the wheel can spin freely Otheexpertta.com > 33% Part (a) If the drive chain exerts a force of 2500 N at a radius of 4.8 cm, what is the angular acceleration of the wheel. in radians per square second? Grade Summary Deductions Potential 0% 100% sin0 tan0 coso cta s acos atan) acotan)sinh0 cosh tan cotanh0 Submissions Attempts remaining (2% per attempt) 10 detailed view 0 Degrees ORadians Submit Hant I give upt deduction per hint Huts remaining Feedback: 0% deduction per fedback Hints: 0 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 82 rad/s?Explanation / Answer
According to the given problem,
(a) The moment of inertia of an annular ring is
I = m*(R12 + R22) / 2
So for the wheel,
I = 12*(0.292 + 0.312) / 2 = 1.0812 kg.m^2
And Torque,
T = 2500N * 0.048m = 120 N-m
Thus, the Angular accelaration,
= T / I = 120 / 1.0812 rad/s2 = 110.987 = 111 rad/s2
(b) a = r* = 0.31 * 111 rad/s2 = 34.41 m/s2
(c) t = / = 82 / 111 = 0.74 s