• Support PF! Buy your school textbooks, materials and every day products Here!

Energy, momentum and centre of mass

  • Thread starter erisedk
  • Start date
  • #1
374
7

Homework Statement


Two spheres, each of mass m and radius r are tied with a spring of spring constant k. The spring is stretched by a certain amount and the system is kept on a rough surface where friction is sufficient to prevent sliding. If one sphere is solid and the other is hollow, then for the spring-sphere system-
(A) Conservation of momentum is valid
(B) Conservation of energy is valid
(C) Displacement of centre of mass is not zero
(D) Velocity of centre of mass is zero

Multiple answers can be correct.

Homework Equations




The Attempt at a Solution


Since this is a rough surface with no sliding, friction will not act on the system because there will be pure rolling. Since net external force is zero, I think (A), (B) should hold.
Since net ext. force is zero, acceleration of centre of mass is also zero. Since velocity of COM before the system was released from rest was zero, and acc. of COM is zero, (D) should hold. And if velocity of COM is zero, (C) should also be true. But my answer is incorrect.
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,050
3,024
How do you conclude that the net external force is zero ?
 
  • #3
254
8
It would be helpful to know where and how the spring is attached to the spheres. If the direction of the force due to the spring is radial in reference to the spheres, then there should be no torque, thus rolling should not occur. (This is based on the first sentence of your attempt, I don't know if I understand your statement o0)).
 
  • #4
BvU
Science Advisor
Homework Helper
2019 Award
13,050
3,024
Second castro: I take it for granted from the problem formulation (or better: the solution attempt) that the spheres can roll towards and away from each other.
 
  • #5
374
7
Don't go by my solution attempt. I may be extremely wrong.
I view this as a spring-mass system in which there is movement of the masses due to stretching of the spring, however, instead of the masses sliding back and forth, they are rolling.
 
  • #6
BvU
Science Advisor
Homework Helper
2019 Award
13,050
3,024
Rolling around a horizontal axis that is perpendicular to the spring axis ?
 
  • #7
374
7
Something that looks like this--
 

Attachments

  • #8
BvU
Science Advisor
Homework Helper
2019 Award
13,050
3,024
Anticlimactical and utterly unhelpful :)
But, never mind: assume the balls can't slide and can only roll. Driving force is the spring force that is equal and opposite for the two balls. Which one would accelerate more than the other ? Or would they have the same |a| (in the latter case I would say no to (c) and yes to (d) ) ?
 
  • #9
374
7
The solid sphere would accelerate more than the other one because it has a smaller moment of inertia. So, that rules out D. C should be true. B is definitely true. What about A (conservation of momentum)?
 
  • #10
BvU
Science Advisor
Homework Helper
2019 Award
13,050
3,024
How can the accelerations be different if the spring force is equal and opposite for the two spheres ?
And: doesn't (c) = true mean something for (a) ?
 
  • #11
jbriggs444
Science Advisor
Homework Helper
2019 Award
8,347
3,186
How can the accelerations be different if the spring force is equal and opposite for the two spheres ?
Because the "f" in f=ma is net force. The force from the spring is not the only force in play here.
 
  • #12
374
7
Yeah, there's the friction that will provide a torque. This will result in a larger angular acceleration for the solid sphere cos it has a smaller MOI, which will lead to a larger linear acceleration (as a= rα for rolling).
Will momentum be conserved?
 
  • #13
ehild
Homework Helper
15,427
1,827
Remember how the acceleration of the CM is related to the time derivative of the momentum.
 
  • #14
374
7
F(ext) = dP/dt.
The external force according to me should be zero as the same frictional force (and the same spring force) acts on both the bodies. So momentum should be conserved. But (A) is incorrect.
 
  • #15
ehild
Homework Helper
15,427
1,827
F(ext) = dP/dt.
The external force according to me should be zero as the same frictional force (and the same spring force) acts on both the bodies. So momentum should be conserved. But (A) is incorrect.
The spring force is the same, but why do you think the frictional forces are also the same. Remember, it is static friction in case of rolling.
 
  • #16
BvU
Science Advisor
Homework Helper
2019 Award
13,050
3,024
Because the "f" in f=ma is net force. The force from the spring is not the only force in play here.
Yes, thank you. ;) The question was intended to get Erise thinking in that direction....

Also somewhat spoiling is that erise already seems to know that (a) is false
 
  • #17
374
7
I don't know why the frictional forces should be different.
 
  • #18
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,786
5,050
I don't know why the frictional forces should be different.
Step through these in turn:
Is the spring force the same on each?
Is the initial torque about their points of contact with the surface the same on each?
Is the initial angular acceleration the same for each?
Is the initial horizontal acceleration the same for each?
Is the net horizontal force the same for each?
 
  • #19
ehild
Homework Helper
15,427
1,827
I don't know why the frictional forces should be different.
The balls roll. In case of rolling, the velocity of the CM is VCM=ωR where ω is the angular velocity and R is the radius of the sphere. The same holds for the accelerations: a=Rα.
The friction f creates torque τ=Rf round the centre of the ball, Rf = I α. Because of rolling, a=Rα=f R2/I.
For the acceleration of the CM, you get ma=F(spring)-f . Eliminate f, and find the expression for the accelerations of both balls. The accelerations are proportional to the spring force, but depend also on the moments of inertia, which are not the same for both balls. The magnitude of accelerations are different: The centre of mass will accelerate.
 
  • #20
374
7
Oh ok! I get it now, thanks everybody :)
 

Related Threads on Energy, momentum and centre of mass

Replies
7
Views
5K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
14
Views
6K
Replies
16
Views
8K
  • Last Post
Replies
1
Views
700
  • Last Post
Replies
0
Views
4K
Replies
7
Views
3K
Replies
15
Views
4K
Replies
1
Views
4K
  • Last Post
Replies
2
Views
14K
Top