Engineering Zero Velocity & Acceleration of Point of Contact: Problem Explanation

[Pipsqueakalchemist]

Homework Statement

Question below

Relevant Equations

a_A = a_B + α X r_BA - w^2*r_BA (relative acceleration)

For this question the ball is rolling without slipping so that means the velocity of the point of contact is zero. Does that also apply to the acceleration of the point of contact? Because that’s what I assumed and I applied the relative acceleration formula above and use the starting point to be the point of contact and ending point to be point C. Then I just plugged everything in but I got the wrong answer. Can someone explain where I went wrong?

 

[PeroK]

If you throw a ball up it has zero velocity at the highest point, but the acceleration at that point is $g$.

 

[Pipsqueakalchemist]

If you throw a ball up it has zero velocity at the highest point, but the acceleration at that point is $g$.

But then I’m confused how the solution found a_o.

 

[PeroK]

But then I’m confused how the solution found a_o.

The disc is accelerating. It's not hard to relate linear acceleration to the given angular acceleration, which is what the solution does.

 

[Pipsqueakalchemist]

So the solution uses the angular acceleration of the disk and the radius from the point of contact to the centre of mass. But then why is doesn’t it include the acceleration of the point of contact. I’m not really sure how this makes sense because when the solution found a_o they used linear acceleration = angular*radius where the radius starts at the centre of rotation and isn’t the centre of rotation suppose to have zero velocity and speed. I guess I’m confused how the solution found a_o without knowing the acceleration of the point of contact.

 

[PeroK]

So the solution uses the angular acceleration of the disk and the radius from the point of contact to the centre of mass. But then why is doesn’t it include the acceleration of the point of contact. I’m not really sure how this makes sense because when the solution found a_o they used linear acceleration = angular*radius where the radius starts at the centre of rotation and isn’t the centre of rotation suppose to have zero velocity and speed. I guess I’m confused how the solution found a_o without knowing the acceleration of the point of contact.

You can calculate the acceleration of the point of contact if you want, but it's not relevant to finding the acceleration of the point C.

 

[Pipsqueakalchemist]

How would you do it then? Is the acceleration at the point of contact restricted only in the horizontal direction bc I could understand doing this question without knowing the magnitude of the point of contact,centre of mass, and point C but if we knew they’re directions.

 

[Pipsqueakalchemist]

Ok so I did some searching and the formula the solution used was ao = alpha x radius. So the equation relates the acceleration of the centre of mass with the wheels angular acceleration. I think my main confusion was that I thought the radius had to be starting from an point with no acceleration. For example if you want to find velocity of the centre of mass you would multiply the angular velocity with the radius starting from the centre of rotation which is the point of contact which has zero velocity. I thought that had to apply to acceleration as well.

 

[PeroK]

I think my main confusion was that I thought the radius had to be starting from an point with no acceleration. For example if you want to find velocity of the centre of mass you would multiply the angular velocity with the radius starting from the centre of rotation which is the point of contact which has zero velocity.

I don't understand why there would be a point on an accelerating rigid body with zero acceleration. There might be, but in general all points on the body are accelerating.

If a rigid body is rotating, then any point can be chosen as the centre of rotation. The most natural is usually the centre of mass. It's not clear why you would choose the point of contact in this case. Also, the point of contact is not a fixed point on the body, so that adds its own complications. Using the centre of mass must be much simpler conceptually and mathematically in this case.