Question

(11%) Problem 9: Consider the 17 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 0.32 m. The motorcycle is on its center stand, so that the wheel can spin freely. Otheexpertta.com A 33% Part (a) If the drive chain exerts a force of 2200 N at a radius of 5.05 cm, what is the angular acceleration of the wheel, in radians per square second? A 33% Part (b) What is the tangential acceleration, in meters per square second, of a point on the outer edge of the tire? 339 Part (c) How long, in seconds, starting from rest, does it take to reach an angular velocity of 77.5 rad/s? Grade Summary 1 = 1 Dedactions P 10

Answer 1

m = 11 kg = 0. 27 m. Yout= 0.32 m moment of inertia of a ring I= 2 m (vint out) I = 1x11 kg x [(0.27m)* +(0:32 m)?] I = 0.96415 kg. 12 a) F= 2200 N R = 5.05 cm = 0.0505 m. angular acceleration & HR 2= 2200 NX 0.050s m. o.gtuis kg.m2 &= 115.231 radla 6) Tangental acceteration is a = Yout a=0.32 m x 115-231 rade a= 36.876 misal c) w = 77.5 rady time to w W=Wotat w on two t = 97.5 rad's 15. 231 sodyga t= 0.673 S