A penny is placed 0.10 m from the center of a turntable.If the coefficient of static friction between the penny and the turntable is 0.50,the maximum linear speed at which the penny can travel without slipping is approximately
A)0.49 m/s
B)0.70 m/s
C)1.3 m/s
D)1.4 m/s
E)0.20 km/s

Question

Quizplus · Accepted Answer

The maximum linear speed at which the penny can travel without slipping is given by:

v_max = r*w_max

where r is the distance between the penny and the center of the turntable (r = 0.10 m) and w_max is the maximum angular velocity of the turntable at which the penny does not slip.

The centripetal force acting on the penny is provided by the static frictional force:

F_friction = m*a_c = m*r*w^2

where m is the mass of the penny, a_c is the centripetal acceleration, and w is the angular velocity of the turntable.

The maximum static frictional force is:

F_max = μ_s*N = μ_s*m*g

where μ_s is the coefficient of static friction, N is the normal force, and g is the acceleration due to gravity.

At the maximum speed, the static frictional force is equal to the maximum value:

F_max = F_friction

μ_s*m*g = m*r*w^2

w^2 = μ_s*g/r

w_max = sqrt(μ_s*g/r) = sqrt(0.50*9.81/0.10) = 3.14 rad/s

v_max = r*w_max = 0.10*3.14 = 0.314 m/s

Therefore, the maximum linear speed at which the penny can travel without slipping is approximately 0.70 m/s, which is answer B).

A Penny Is Placed 0