Q.

A disc having radius R is rolling without slipping on a horizontal plane as shown. Centre of the disc has a velocity v and acceleration a as shown.  Speed of point P having coordinates (x, y) is If V=2aR the angle θ between acceleration of the topmost point and the horizontal is

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a

Vx2+y2R

b

Vx2+(y+R)2R

c

vx2+(y−R)2R

d

None

e

0

f

45°

g

tan−1 ⁡(2)

h

tan−1 ⁡12

answer is , .

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Detailed Solution

∵ω=VR,OP=x2+y2origin is ins tantaneous centre of rotation∴VP=ωr=ω(OP)=VRx2+y2∵a=αRac=ω2R=VR2R=V2R=(2a)tan⁡θ=ac2a=1∴θ=45∘
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