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Question

A uniform solid cylinder of radius R=15cm rolls over a horizontal plane passing into an inclined plane forming an angle α=30 with the horizontal (figure shown above). Find the maximum value of the velocity v0 in (m/s) which still permits the cylinder to roll onto the inclined plane section without a jump. The sliding is assumed to be absent.
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Solution

Since the cylinder moves without sliding, the centre of the cylinder rotates about the point O becomes the foot of the instantaneous axis of rotation of the cylinder.
If at any instant during this motion the velocity of the C.M. is v1 when the angle (shown in figure below) is β, we have
mv21R=mgcosβN,
where N is the normal reaction of the edge
or, v21=gRcosβNRm (1)
From the energy conservation law,
12I0v21R212I0v20R2=mgR(1cosβ)
But I0=mR22+mR2=32mR2,
(from the parallel axis theorem)
Thus, v21=v20+43gR(1cosβ) (2)
From (1) and (2)
v20=gR3(7cosβ4)NRm
The angle β in this equation is clearly smaller than or equal to α so putting β=α we get
v20=gR3(7cosα4)N0Rm
where, N0 is the corresponding reaction. Note that NN0. No
jumping occurs during this turning if N0>0. Hence, v_0 must be less than
vmax=gR3(7cosα4)=1m/s
228299_147397_ans.png

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