Confused on whether this counts as an external torque

  • #1
laser
81
11
Homework Statement
Posted in description
Relevant Equations
Tension in this case is Mg, which is equal to centripetal force, mv^2/r
1701686067504.png

For part (d), there is the formula a = v^2/r I can use. Note that Mg = mv^2/r, we have two unknowns, v and r. I can solve this if conservation of angular momentum is true, i.e. mvr = constant. I am not convinced I can use this however, because is increasing M torque?

My idea is that it is an external torque, because by the right hand rule, torque points along the axis which M is acting on.

How can I solve this problem?

(This problem comes from a past exam paper, which the physics department will not provide help with)
 
Physics news on Phys.org
  • #2
laser said:
because is increasing M torque?
What torque? The force from the string on the mass m is radial.
 
  • Like
Likes laser
  • #3
Damn, you're right. I kind of forgot their configuration. Thanks!
 
  • #4
Orodruin said:
What torque? The force from the string on the mass m is radial.
One more doubt: The conservation of angular momentum is valid if there are no external torques. Even though this torque is not interfering with the motion of the object rotating in a circle, there is still an external torque being applied to the system (radially). Why can we still say the conservation of angular momentum is constant? Is it only constant in the plane that m rotates in?
 
  • #5
laser said:
Even though this torque is not interfering with the motion of the object rotating in a circle, there is still an external torque being applied to the system (radially).
There is no torque
.
Consider a force F and some point P (which can be anywhere).

Let d be the perpendicular (shortest) distance between F's line-of-action and P.

The magnitude of the torque (about P) is Fd.

But if P lies on the force's line-of-action, then d=0. In this case, the torque about P is Fd = Fx0 = 0.

In the current question, the tension acting on the mass is radial - it acts through the centre of rotation: so d=0 making the torque (about the centre of rotation) zero.

Consider trying to open a screw-top cap on bottle using a radial force! There is zero torque about the centre of rotation and the cap doesn't turn. You need a tangential force (or component of force) to create a torque.
 
  • Like
Likes laser

Similar threads

  • Introductory Physics Homework Help
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
171
  • Introductory Physics Homework Help
Replies
5
Views
521
  • Introductory Physics Homework Help
Replies
7
Views
252
  • Introductory Physics Homework Help
Replies
10
Views
718
  • Introductory Physics Homework Help
10
Replies
335
Views
8K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
152
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
2
Replies
54
Views
2K
Back
Top