Solid uniform ball of mass M rolls down 45° incline without slipping. What is the force of friction?

Question

acting on the ball?

Answer 1

Frictional force
= mg sin θ * [ k^2 / (k^2 + R^2) ],
where k = radius of gyration = R √(2/5) for solid sphere

=> Frictional force
= Mg (1/√2) * [ (2/5)R^2 / [ (2/5)R^2 + R^2 ] ]
= (Mg √2) / 7

Answer 2

The frictional force F is the source of torque in I alpha = F r; where r is the ball's radius, I = 2/5 Mr^2 which is the angular momentum for a solid sphere (aka ball), and alpha = a/r is the angular acceleration where the center of the ball is accelerating a linearly.

Thus, F = I alpha/r = 2/5 M r^2(a/r)//r = 2/5 Ma. From the net force acting on the ball f = Ma = Mg sin(theta) - 2/5 Ma; a = g sin(theta) - 2/5 a and a = 5/7 g sin(theta); where theta = 45 deg ramp incline. Thus, F = 2/5 Ma = 2/5 M(5/7 g sin(theta)) = 2/7 M g sin(45) = 1.98 M if I did my arithmetic right.

Answer 3

Since its not slipping all of your friction with the plane is due to static friction. Dont get bogged down with how the math works, think physically whats going on.

If the incline was at 90 degress the friction would be 0 because there is no weight pushing down on the incline (Friction proportinal to perpendicular weight). So you know to use the cosine fuction because at 90 degrees it will go to 0. So your equation is Mg cos(45). This is the force of the ball going parellel to the incline. But since there is no slipping the Friction force must balance this..so its Mg cos(45)..and the direction is up the plane.

Answer 4

linear acceleration = g/√2 - f/M
angular acceleration = g/r√2 - f/Mr
Torque on ball = f*r = 2/5 Mr^2 *[ g/r√2 - f/Mr]
7/5 f = 2/5 Mg/√2
f = 2/7 Mg/√2
= Mg*(√2)/7

Answer 5

Use conservation of energy: 1/2*m*v^2_i + 1/2*I*w^2_i + mgh_i = 1/2*m*v^2_f + 1/2*I*w^2_f + mgh_f 0 + 0 + mgh_i = 1/2*m*v^2_f + 1/2*I*w^2_f + 0 <-------- substitute v^2 for w*r <---------w^2*r^2 mgh_i = 1/2*m*w^2*r^2 + 1/2*I*w^2 2(mgh_i) = m*w^2*r^2 + I*w^2 2(mgh_i)/(m*r^2 + I) = w^2 2(137.2)/(8.092) = w^2 5.82 rad/s = w

Answer 6

mgsin(theta)
M(9.8)sin(45) = F

Answer 7

(sqrt(2)/2)(g)(M)