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Question

A uniform hollow cylinder of mass 6.5 kg, inner radius 0.13 m and outer radius 0.25 m approaches a flat ramp and rolls up the ramp without slipping. The ramp is inclined at 20o. How far along the ramp does the cylinder roll before stopping, if its initial forward speed is 12 m/s ?

(Given: sin20o =0.342)

43885_d23b45a1664a45ad9094ca7760547985.png

A
25 m
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B
15 m
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C
35 m
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D
70 m
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Solution

The correct option is C 35 m
Initially, there are translation and rotational kinetic energies. When the cylinder stops momentarily at the top of the ramp , the kinetic energy has totally transformed to gravitation potential energy.
12Mv2+12Iω2=Mgy
For the cylinder
I=12M(R21+R22)
The speed of the cylinder is the tangential speed of the wheel at the outer radiusR2
ω=vR2
12Mv2+12[12M(R21+R22)](vR2)2=Mgy
y=v2g[12+14(R21+R22R22)]
=(12)210[12+14{(0.13)2+(0.25)2(0.25)2}]
y=12m= cylinder's maximum vertical height
Distance rolled by cylinder along ramp= L =ysin20035m

118513_43885_ans_6d7180a6880c44e4b8b9a64fac0ed5ad.png

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