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Question

A wheel (to be considered as a ring) of mass m and radius R rolls without sliding on a horizontal surface with constant velocity v. It encounters a step of height R2 at which it ascends without sliding. Choose the correct statement(s).


A
The angular velocity of the ring just after it comes in contact with the step is 3v4R.
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B
The normal reaction due to the step on the wheel just after the impact is mg29mv26R.
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C
The normal reaction due to the step on the wheel increases as the wheel ascends.
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D
The normal reaction due to the step on the wheel stays constant as the wheel ascends.
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Solution

The correct option is C The normal reaction due to the step on the wheel increases as the wheel ascends.

Angular momentum about IAOR remains conserved. [no net external torque]

After hitting the step, point O behaves as the instantaneous axis of rotation. Let ωi be the initial angular velocity about centre and ω be the angular velocity about point O immediately after collision.

Then,
Li=Lf

Icωi+mvir=IOω


ω=Icωi+mvirIO

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢Ic=mR2, ωi=vRr=R2, vi=vIO=mR2+mR2 =2mR2⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

ω=mR2(vR)+mv(R2)2mR2

ω=3v4R (1)

Now, just after impact
mgcos60N=mv2cR [from FBD]

N=mg2m(Rω)2R (2) [vc=Rω]

Using (1) in (2),

N=mg29mv216R

As the wheel ascends, θ (which was initially 60) decreases and vc decreases [from conservation of mechanical energy]

N=mgcosθmv2cR

Thus, N increases.

So, options (a),(b) and (c) are correct.
Why this question?
This is a multi-conceptual question that involves the concepts of instantaneous axis of rotation, angular momentum conservation and Newton's laws in circular motion.

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