Recent content by DragonZero

So would that be: T - Mg(coefficient of kinetic friction) = mg - T

Homework Statement In the system shown below, there is a block of mass M = 4.4 kg resting on a horizontal ledge. The coefficient of kinetic friction between the ledge and the block is 0.25. The block is attached to a string that passes over a pulley, and the other end of the string is...

Thank you.

torque = r * f = r * m*g*sin(angle) = 0.19 * 8 * 9.8 * sin(24) = 6.05 inertia about contact point = inertia center of mass + mass * distance^2 = 0.11552 + 8 * 0.19^2 = 0.40432

How should i go about calculating torque? I know it's inertia * alpha, but since I don't yet have a linear acceleration I can't use alpha = a/R. I know torque also equals radius * force, so for that force would I plug in m*g*sin(angle)?

Homework Statement A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle θ = 24°. The sphere has mass M = 8.0 kg and radius R = 0.19 m. The coefficient of static friction between the sphere and the plane is μ = 0.64. What is the...

It worked! Thank you very much!

43 = I * (3.19 + 0.357) I = 12.11 T_f = 12.11 * 0.358 = 4.33

For speeding up: 43 - T_f = I * 3.19 For slowing down: T_f = I * 0.358 Eqeuation: 43 - I * 0.358 = I * 3.19

For speeding up: 43 - T_f = I * 3.19 For slowing down: 43 = I * 0.358 So what do I do next? Is it (I * 3.19) + T_f = I * 0.358

Ok, so for the angular velocity I did (610 * 2 pi)/60 for radians/sec to get a speed of 63.879. And I have to use Newton's second law, torque = inertia * angular acceleration. angular velocity = angular acceleration * time For the speeding up: 63.879 / 20 = angular acceleration =...

Homework Statement A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +43 N·m is applied to the wheel for 20 s, giving the wheel an angular velocity of +610 rev/min. The external torque is then removed, and the wheel comes to...