A rolling ball accelerating down an incline

AI Thread Summary
In the discussion about a ball rolling down an incline, confusion arises when static friction equals the gravitational component along the plane, resulting in zero net force but non-zero net torque. It is clarified that static friction is not a fixed value; it adjusts to prevent slipping, up to a maximum determined by the coefficient of static friction. If the gravitational force exceeds this maximum, the ball will slip, transitioning to kinetic friction. The analysis concludes that if friction and gravitational forces are equal, it leads to contradictions in Newton's laws, indicating an unstable equilibrium. Ultimately, the ball will begin to move despite the initial conditions suggesting otherwise.
Mr Sun
Messages
2
Reaction score
0

Homework Statement


summer_14_sketches_key10.jpg

source:http://www.wired.com/2014/07/a-rolling-object-accelerating-down-an-incline/

For a ball rolling on an incline, I know how to calculate the acceleration. However, I am quite confused about a situation. What if static friction acting on the ball is equal to the component of gravity along the plane, which means, net force acting on the ball is zero. But net torque of the ball is not zero, which means the ball is supposed to rotate. Under this situation, either the ball rolling down or not is quite confusing. Considering net force, it should not have translation. but considering net torque, it should rotate, and then roll down.

What will happen ? Is there any mistake on my analysis ? Please help! Thanks a lot!
 
Physics news on Phys.org
There are 2 ways to solve this problem.
(1) You can take torque about the point of contact with the plane (using the parallel axis theorem).
In this case the frictional force does not enter into the solution.
(2) You can take torque about the center of mass with the frictional force providing the torque.
Here you also have to consider the net translational force on the object and then
you can eliminate the frictional force from the resulting equations.
 
But is there any problem in my understanding? Why will a ball translate while net force is zero?
 
Mr Sun said:

Homework Statement



source:http://www.wired.com/2014/07/a-rolling-object-accelerating-down-an-incline/

For a ball rolling on an incline, I know how to calculate the acceleration. However, I am quite confused about a situation. What if static friction acting on the ball is equal to the component of gravity along the plane, which means, net force acting on the ball is zero. But net torque of the ball is not zero, which means the ball is supposed to rotate. Under this situation, either the ball rolling down or not is quite confusing. Considering net force, it should not have translation. but considering net torque, it should rotate, and then roll down.
Static friction is not a force of definite value. It gets a value what is necessary for rolling. That force can not be greater than μsFN. If the component of gravity is greater than this maximum force of static friction, the friction becomes kinetic and the ball will slip. But the force of static friction can be anything smaller then the maximum value.
 
If the frictional force equaled the component of the gravitational force along the plane the net translational force would be zero.
This contradicts the fact that frictional force supplies the torque that causes the ball to roll.
You can verify this by calculating the actual forces (frictional and translational) and show that they are not equal.
If they are equal then Newton's laws are invalid.
Obviously, the system is in unstable equilibrium, and the ball is going to move.
Try balancing a pencil on its point and say that the pencil will not fall because the frictional
force on the point of the pencil balances the gravitational force that acts on the center of mass of
the pencil so the center of mass of the pencil will not move.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top