A disc of mass M and radius R rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is v, the height to which the disc will rise will be:

Question

A disc of mass M and radius R rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is v, the height to which the disc will rise will be: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/212e97bf1732cfe42b29f76c1a783224-.png" /> (A) (3 v2/2 g) (B) (3 v2/4 g) (C) (v2/4 g) (D) (v2/2 g)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/08cc3121d9e67440609f63485b9b644f-.png" /> (1/2) mv 2+(1/2) I ω2= mgh (1/2) m v2+(1/2) (M R2/2) (v2/R2)=m g h (3/4) m v2=mgh h=(3 v2/4g)

Q. A disc of mass $M$ and radius $R$ rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is $v$ , the height to which the disc will rise will be:

$\frac{3 v ^{2}}{2 g}$

$\frac{3 v ^{2}}{4 g}$

$\frac{v ^{2}}{4 g}$

$\frac{v ^{2}}{2 g}$

$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = m g h$
$\frac{1}{2} m v^{2} + \frac{1}{2} \frac{M R ^{2}}{2} \frac{v ^{2}}{R ^{2}} = m g h$
$\frac{3}{4} m v^{2} = m g h$
$h = \frac{3 v ^{2}}{4 g}$

Q. A disc of mass M and radius R rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is v, the height to which the disc will rise will be:

2g3v2​

4g3v2​

4gv2​

2gv2​