An object of radius R and mass M is rolling horizontally without slipping with speed v. It then rolls up the hill to a maximum height is h=(3 v2/4 g). The moment of inertia of the object is ( g= acceleration due to gravity):

Question

An object of radius R and mass M is rolling horizontally without slipping with speed v. It then rolls up the hill to a maximum height is h=(3 v2/4 g). The moment of inertia of the object is ( g= acceleration due to gravity): (A) (2/5)MR2 (B) (M R2/2) (C) MR2 (D) (3/2)MR2

Tardigrade Admin · Accepted Answer

Loss in KE= Gain in PE (1/2)mv2(1 + (K2/R2))=mgH (v2/2)(1 + (K2/R2))=g((3 v2/4 g)) (K2/R2)=(1/2) I=MK2=(M R2/2)

Q. An object of radius $R$ and mass $M$ is rolling horizontally without slipping with speed $v .$ It then rolls up the hill to a maximum height is $h = \frac{3 v ^{2}}{4 g} .$ The moment of inertia of the object is
( $g =$ acceleration due to gravity):

$\frac{2}{5} M R^{2}$

$\frac{M R ^{2}}{2}$

$M R^{2}$

$\frac{3}{2} M R^{2}$

Loss in $K E =$ Gain in $PE$
$\frac{1}{2} m v^{2} (1 + \frac{K ^{2}}{R ^{2}}) = m g H$
$\frac{v ^{2}}{2} (1 + \frac{K ^{2}}{R ^{2}}) = g (\frac{3 v ^{2}}{4 g})$
$\frac{K ^{2}}{R ^{2}} = \frac{1}{2}$
$I = M K^{2} = \frac{M R ^{2}}{2}$

Q. An object of radius R and mass M is rolling horizontally without slipping with speed v. It then rolls up the hill to a maximum height is h=4g3v2​. The moment of inertia of the object is ( g= acceleration due to gravity):

52​MR2

2MR2​

MR2

23​MR2