Stupid Question...sorry Wasn't thinking. Ok so I have the Linear Acceleration. But how do I figure out when the ball actually starts to roll and stop sliding?
Ok so I got that angular acceleration during sliding is 42.3 rad/s by the fact that, T = I\alpha and also equal to the equation above.
I realize that when the ball stops slipping that the Acceleration of the Center of mass = \alpha * R and that all those equations start to apply where...
[SOLVED] Rolling Forces
Homework Statement
A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides on the lane with initial speed vcom,0 = 6.0 m/s and initial angular speed 0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.19. The...
[SOLVED] Angular Rotation
Homework Statement
A cockroach of mass m lies on the rim of a uniform disk of mass 7.50 m that can rotate freely about its center like a merry-go-round. Initially the cockroach and disk rotate together with an angular velocity of 0.250 rad/s. Then the cockroach...
In the figure below, a constant horizontal force Fapp of magnitude 15 N is applied to a wheel of mass 14 kg and radius 0.40 m. The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude 0.43 m/s2.
(a) In unit-vector notation, what is the...
Force F = (9.0 N) i hat + (-4.0 N) k hat acts on a pebble with position vector r = (3.00 m) j hat + (-1.0 m) k hat, relative to the origin.(a) What is the resulting torque acting on the pebble about the origin?
( N·m) i hat + ( N·m) j hat + ( N·m) k hat
(b) What is the resulting torque...
Shouldn't both blocks be accelerating in the positive x axis?
Or are you saying that the equation is:
F - Fs = ma ??
shouldn't the frictional force be positive since it is pointing towards the acceleration, similar to how a car works?
Ok now I think I'm getting a little closer and yet still further away at the same time.
So the equation for the top block is:
Fs = Frictional Force
F = applied force
m1 = mass of top block
m2 = mass of bottom block
Fs - F = ma
while the equation for the bottom block is:
F = ma...
Ok so the acceleration of both blocks is equal. Also the only force acting on the top block is the static frictional force. Therefore:
The force of the bigger block is
F = umg = ma
F = 9.016 = ma
a = 2.254m/s2
Then since the acceleration is equal I plug the acceleration into...
[SOLVED] Force Problem
A 4.0kg block is put on top of a 5.0 kg block. The coefficient of static friction between the two blocks is u = 0.23 . The assembly of blocks is placed on a horizontal, frictionless table. What is the magnitude of the maximum horizontal force F that can be applied to...