# Explain the phenomenon of decrement of angular momentum

• AyushNaman
AyushNaman
Homework Statement
How can we explain the phenomenon of the decrement of angular momentum of a cubical block (let's say) that's moving on a horizontal rough surface, purely in the inertial frame of reference?
Relevant Equations
$$L=mvr$$
I tried to work out the net resultant postion of the normal force but could only come at a conclusion that normal force and mg, both pass through C.O.M(torques were considered about the edge of block).

AyushNaman said:
Homework Statement: How can we explain the phenomenon of the decrement of angular momentum of a cubical block (let's say) that's moving on a horizontal rough surface, purely in the inertial frame of reference?
Relevant Equations: $$L=mvr$$

I tried to work out the net resultant postion of the normal force but could only come at a conclusion that normal force and mg, both pass through C.O.M(torques were considered about the edge of block).
By "decrement" do you mean "decrease"? And what inertial frame are you referring to? The frame of the surface (assuming the surface is attached to a much larger "fixed" body to make it inertial? If the block is experiencing a deceleration with respect to the surface due to friction, a reference frame attached to the block will not be inertial.

Could you maybe upload a sketch of the problem? (use "Attach files" to upload a PDF or JPEG file)

AyushNaman said:
Homework Statement: How can we explain the phenomenon of the decrement of angular momentum
About what axis? If the axis is at ground level, normal to the trajectory of the block, then the angular momentum of the block about the axis is, mvr, where r is the height of the mass centre. So as v decreases, mvr decreases.
I assume your question is "what torque causes this decrease?"

There are three forces on the block: gravity, the normal force and friction. Since the friction acts at ground level, it creates a torque tending to tip the block forward. That shifts the distribution of the normal force towards the front of the block. (So long as its average position stays within the footprint of the block it will not tip.) But gravity continues to act through the mass centre, so the N, mg pair produce a torque opposing the angular momentum of the block.

AyushNaman said:
Homework Statement: How can we explain the phenomenon of the decrement of angular momentum of a cubical block (let's say) that's moving on a horizontal rough surface, purely in the inertial frame of reference?
Relevant Equations: $$L=mvr$$

I tried to work out the net resultant postion of the normal force but could only come at a conclusion that normal force and mg, both pass through C.O.M(torques were considered about the edge of block).
The angular momentum is a vector. It's magnitude is not mvr unless the two vectors ##\vec{r} ## and ##\vec{v} ## are perpendiuclar to each other. Is this the case in your setup?

vanhees71

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