Recent content by Nanuven

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...

Ok, I got it. Thanks everyone!

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...