If a sphere rolling on an inclined plane with velocity v without slipping, the vertical height of the incline in terms of velocity will be:

Question

If a sphere rolling on an inclined plane with velocity v without slipping, the vertical height of the incline in terms of velocity will be: (A)  (7 upsilon /10g)  (B)  (7 upsilon 2/10g)  (C)  (2 upsilon 2/5g)  (D)  (2 upsilon /5g)

Tardigrade Admin · Accepted Answer

The velocity of solid sphere on the bottom of inclined plane is

v = [\frac{2 g h}{( 1 + I / M R ^{2} )}]

where,

I =

moment of inertia of sphere,

M =

mass of sphere,
and

R =

radius of sphere
The moment of inertia of solid sphere about its diameter

I = \frac{2}{5} M R^{2}

∴ v = [\frac{2 g h}{( 1 + \frac{2}{5} )}] = (\frac{10}{7} g h)

∴ h = \frac{7 v ^{2}}{10 g}

Q. If a sphere rolling on an inclined plane with velocity $v$ without slipping, the vertical height of the incline in terms of velocity will be:

$\frac{7 υ}{10 g}$

$\frac{7 υ ^{2}}{10 g}$

$\frac{2 υ ^{2}}{5 g}$

$\frac{2 υ}{5 g}$

Q. If a sphere rolling on an inclined plane with velocity v without slipping, the vertical height of the incline in terms of velocity will be:

10g7υ​

10g7υ2​

5g2υ2​

5g2υ​

The velocity of solid sphere on the bottom of inclined plane is v=[(1+I/MR2)2gh​]​ where, I= moment of inertia of sphere, M= mass of sphere, and R= radius of sphere The moment of inertia of solid sphere about its diameter I=52​MR2 ∴v=[(1+52​)2gh​]​=(710​gh)​ ∴h=10g7v2​

Q. If a sphere rolling on an inclined plane with velocity $v$ without slipping, the vertical height of the incline in terms of velocity will be:

$\frac{7 υ}{10 g}$

$\frac{7 υ ^{2}}{10 g}$

$\frac{2 υ ^{2}}{5 g}$

$\frac{2 υ}{5 g}$