A round uniform body of radius R, mass M and moment of inertia I, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration

Question

A round uniform body of radius R, mass M and moment of inertia I, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration (A) (g sin θ/1+(I)MR2) (B) (g sin θ/1+(MR2)I) (C) (g sin θ/1-(I)MR2) (D) (g sin θ/1-(MR2)I)

Tardigrade Admin · Accepted Answer

Mg sin θ-f=Ma fR=I (a/R) ⇒ a=(g sin θ/(1+(I)MR2))

Q. A round uniform body of radius $R$ , mass $M$ and moment of inertia $I$ , rolls down (without slipping) an inclined plane making an angle $θ$ with the horizontal. Then its acceleration

$\frac{g s in θ}{1 + \frac{I}{M R ^{2}}}$

$\frac{g s in θ}{1 + \frac{M R ^{2}}{I}}$

$\frac{g s in θ}{1 - \frac{I}{M R ^{2}}}$

$\frac{g s in θ}{1 - \frac{M R ^{2}}{I}}$

$M g s in θ - f = M a$
$f R = I \frac{a}{R}$
$\Rightarrow a = \frac{g s in θ}{( 1 + \frac{I}{M R ^{2}} )}$

Q. A round uniform body of radius R, mass M and moment of inertia I, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration

1+MR2I​gsinθ​

1+IMR2​gsinθ​

1−MR2I​gsinθ​

1−IMR2​gsinθ​