Recent content by prettydumbguy

Hmm, so if I have theta=0, AB and BD are both going straight up and D is at its highest point, what happens then? The normal acceleration in AB doesn't change, but then the acceleration of BD would be (alpha x rBD) + w x (w x rBD). So the tangential acceleration would be k x j = i. Then the...

I saw this problem in my dynamics class: http://d2vlcm61l7u1fs.cloudfront.net/media%2Fa22%2Fa2261e7a-0e85-45e7-8b63-f23157d623bc%2Fphp37ql1a.png I know how to solve it, but I don't understand why A(bd) has both a normal and tangential component to acceleration. If it is rotating at a constant...

No angles were given, so I assume that they are vertical. The question comes after two much more in depth SE problems, hence my confusion. Thanks!

Homework Statement A man stands on a board of negligible mass with a length of 10m is supported by 2 cables, one on the left on one on the right, with a tension of 300N and 200N respectively. How much does the man weigh in Newtons?Homework Equations F=ma Torque= F*lever arm The Attempt at a...

Is the tangential acceleration zero though, or is it just the net acceleration that is zero due to the opposite acceleration cause by friction? It's actually part of a larger question that stumped me, where my tangential acceleration at the point of contact was not equal to the translational...

I can't see how they would be different. When an object rolls without slipping, the linear distance it travels is equal to the angular distance multiplied by the radius: a ball with radius R that rotated by one radian will have translated a linear distance equal to R. So it makes sense to me...

Indeed I misspoke, I did mean tangential acceleration as the product of alpha and the radius.

Homework Statement In rotational motion, the tangential velocity is defined as alpha multiplied the radius, When an object is rolling without slipping, the acceleration of the center of the mass is defined as alpha multiplied by the radius. How, if at all, are these two alphas different...

I'm so lost I can't even ask the question clearly.

The force is applied to the center of the wheel in the diagram. 10N moves an object of mass 10kg 0.6m/s^2, or 6N, so there is a force of friction moving in the opposite direction of -4N. 4N*r = Icm(.6/.3) so Icm = 0.6 kgm^2. So in the second question, I use the force of friction acting...

I know, that's my question. Gravity will not cause rotation, only translation. So why is friction not included. Let me use another example. A horizontal force of 10N is applied to a wheel of mass 10 kg and a radius 0.3 m. The wheel moves smoothly on the horizontal surface and the acceleration...

I get that gravity and friction balance so it rotates and translates, but I just don't understand why it is not included. m*g*sin(theta) is the force of gravity acting along the plane (translational force), I don't understand how just adding the radius makes it a rotational force. Let me use...

Oh, I think I get it. So I can view the torque as either a force of friction occurring at the point of contact (axis of rotation) that rotates the center OR a force that occurs at the center that rotates it around the point of contact (axis of rotation), right?

Yes. I don't think I'm asking this question well. If torque comes from the friction between the plane and the sphere, why is there no friction in the equation for torque?

But I didn't have to calculate it, I set m*g*r*sin(theta) = 2/3 M r^2 a/r, and it simplified so m, g, and r all canceled out. My question is just why is m*g*r*sin(theta) equal to torque at all? Doesn't the torque come from the friction between the plane and shell?