Course Contents» » Extra Credit» Notes ·Bookmark einfo Evaluate Communicate Prin
ID: 1508202 • Letter: C
Question
Course Contents» » Extra Credit» Notes ·Bookmark einfo Evaluate Communicate Print Info Print M, a solid cylinder (mass = 1.43 kg, r 0.030 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.610 kg mass, ie, F = 5.978 N. Calculate the angular acceleration of the cylinder. 2.79×102 rad/s^2 You are correct. Your receipt ne tis 168-2969 Previous Tries If instead of the force F an actual mass m =0.610 kg is hung from the string, If instead of the force F an actual mass m 0.610 kg is hung from the string, what is the angular acceleration of the cylinder. 278.9rad/sA2 Hint: In both this part and the first part, the tension in the string induces the torque on the cylinder. In this case, the tension is NOT mg! Use Newton's second law to determine the net force acting on the mass, then solve it for the tension. Then use Newton's second law for rotational motion to determine the angular acceleration of the cylinder. Submit Answer Incorrect. Tries 3/20 Previous Tries How far does m travel downward between 0.630 s and 0.830 s after the motion begins? Submit Answer Tries 0/20 Post Discussion , send FeedbackExplanation / Answer
Torque T = I*alpha
R*T = (0.5*M*R^2)*alpha
mg-T = ma
alpha= a/R
R*T =(0.5*M*R^2)*(a/R)
T = 0.5*M*a
m(g-a) = 0.5*M*a
m*g = (0.5M+m)*a
accelaration a = m*g/(0.5M+m) = (0.61*9.81)/((0.5*1.43)+0.61) = 4.52 m/s^2
angular accelaration alpha = a/R = 4.52/0.03 = 150.67 rad/s^2
distance travelled for t = 0.63 sec is d1 = 0.5*a*t1^2 = 0.5*4.52*0.63^2 = 0.896 m
for t = 0.83 sec ,d2 = 0.5*a*t2^2 = 0.5*4.52*0.83^2 = 1.556 m
then required distance is d2-d1 = 1.556-0.896 =0.66 m