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# Rolling without slipping problems

David explains how to solve problems where an object rolls without slipping. Created by David SantoPietro.

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• At isn't the height 6m? •   No, if you think about it, if that ball has a radius of 2m. So when the ball is touching the ground, it's center of mass will actually still be 2m from the ground. It's true that the center of mass is initially 6m from the ground, but when the ball falls and touches the ground the center of mass is again still 2m from the ground. That means the height will be 4m. If you work the problem where the height is 6m, the ball would have to fall halfway through the floor for the center of mass to be at 0 height.
• I could have sworn that just a couple of videos ago, the moment of inertia equation was I=mr^2, but now in this video it is I=1/2mr^2. Please help, I do not get it. • At energy conservation is used which is only applicable in the absence of non conservative forces. However, isn't static friction required for rolling without slipping? • I have a question regarding this topic but it may not be in the video. Suppose a ball is rolling without slipping on a surface( with friction) at a constant linear velocity.
We know that there is friction which prevents the ball from slipping. Why doesn't this frictional force act as a torque and speed up the ball as well?The force is present. It can act as a torque.
Also consider the case where an external force is tugging the ball along. In this case, my book (Barron's) says that friction provides torque in order to keep up with the linear acceleration.

In the first case, where there's a constant velocity and 0 acceleration, why doesn't friction provide
torque?
Can someone please clarify this to me as soon as possible?
Thanks a lot!!
#APphysicsCMechanics • If the ball is rolling without slipping at a constant velocity, the point of contact has no tendency to slip against the surface and therefore, there is no friction. If the ball were skidding and rolling, there would have been a friction force acting at the point of contact and providing a torque in a direction for increasing the rotational velocity of the ball. This increase in rotational velocity happens only up till the condition V_cm = R.ω is achieved. So friction force will act and will provide a torque only when the ball is slipping against the surface and when there is no external force tugging on the ball like in the second case you mention.
In the second case, as long as there is an external force tugging on the ball, accelerating it, friction force will continue to act so that the ball tries to achieve the condition of rolling without slipping.
• ; At the split second in time v=0 for the tire in contact with the ground. How is it, reference the road surface, the exact opposite point on the tire (180deg from base) is exhibiting a v>0? Surely the finite time snap would make the two points on tire equal in v? Unless the tire is flexible but this seems outside the scope of this problem... • I really don't understand how the velocity of the point at the very bottom is zero when the ball rolls without slipping. Could someone re-explain it, please?
Thanks a bunch! • Haha nice to have brand new videos just before school finals.. :)
As the rolling will take energy from ball speeding up, it will diminish the acceleration, the time for a ball to hit the ground will be longer compared to a box sliding on a no-friction -incline.
How could the exact time be calculated for the ball in question to roll down the incline to the floor (potential-level-0)?
Would it work to assume that as the acceleration would be constant, the average speed would be the mean of initial and final speed. And as average speed times time is distance, we could solve for time.
Would there be another way using the gravitational force's x-component, which would then accelerate both the mass and the rotation inertia? • Nice question. When an object rolls down an inclined plane, its kinetic energy will be
K = Mv²/2 + I.w²/2 , you're probably familiar with the first term already, Mv²/2, but Iw²/2 is the energy aqcuired due to rotation. I is the moment of mass and w is the angular speed. Let's take a ball with uniform density, mass M and radius R, its moment of inertia will be (2/5).MR² (in exams I have taken, this result was usually given). The rotational kinetic energy will then be
M.(R.w)²/5 = Mv²/5, since Rw = v in the described situation.
Therefore, the total kinetic energy will be (7/10)Mv², and conservation of energy yields
-gΔh = (7/10)v²
v = sqrt(-10gΔh/7)
If the inclination angle is a, then velocity's vertical component will be
v.sin a = sqrt(-10gΔh/7).sin a
You can still assume acceleration is constant and, from here, solve it as you described.   