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Question

A uniform solid cylinder of radius R=15 cm rolls over a horizontal plane passing into an inclined plane forming an angle α = 30 with the horizontal. Find the maximum value of the velocity v which still permits the cylinder to roll onto the inclined plane section without a jump. Assume there is no sliding between the cylinder and surfaces.


A
1 m/s
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B
2 m/s
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C
4 m/s
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D

0.5 m/s
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Solution

The correct option is A 1 m/s
At the instant shown in figure, the cylinder rotates about the point O.


Let the velocity of the center of mass of the cylinder be v1 at this instant. As there is no sliding, we can say
N=mgcosβmv21R
where N is normal reaction at the edge O.
or v21=gRcosβNRm....(i)

Change in vertical height of center of mass when the cylinder is at O is R(1cosβ).

From energy conservation, we get
Increase in rotational KE = Decrease in gravitational PE
12Iov21R212Iov20R2 = mgR(1cosβ)....(ii)
Also, Io=12mR2+mR2=32mR2 (from the parallel axis theorem, about IAOR)
Putting values of Io in eq. (ii)
34mR2R2[v21v20]=mgR(1cosβ)
or v21=v20+43gR(1cosβ)..(iii)

Comparing equation (i) & (iii),
gRcosβNRm=v2o+43gR(1cosβ)
or v20=gR3(7cosβ4)NRm

The limitiing value for normal reaction N is N=0, which will occur when β=α.
For all other β values, N>0
we put β=α & N=0
{for these conditions v0=vmax}

v2max=gR3(7cosα4)
vmax=gR3(7cosα4)
Putting α=30 & R=0.15 m
vmax1=1 m/s

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