Question

2kg disc of radius 1m moment of inertia 0.5kgm^2 is released from top of 2m long incline having inclination 30degree. Find it’s velocity just before reaching the ground .??

Answer 1

Given

Mass of the disc `M=2kg`

Radius of the disc `r=1m`

Length of incline `l=2m`

Moment inertia of the disc `I=0.5kgm^2`

Angle of inclination of the incline `theta=30^@`
When the disc reaches at the ground let the linear velocity of the disc be `v` and angular velocity be `omega`. At the ground whole of it's gravitational PE will be transfromed into rotational and translational KE.

So

`1/2Iomega^2+1/2Mv^2=Mglsintheta`

`=>1/2Iv^2/r^2+1/2Mv^2=Mglsintheta`

`=>Iv^2/r^2+Mv^2=2Mglsintheta`

`=>v^2(I/r^2+M)=2Mglsintheta`

`=>v^2(0.5/1^2+2)=2xx2xx9.8xx2xxsin30^@`

`=>v^2xx5/2=2xx2xx9.8xx2xx1/2`

`=>v^2=2xx2xx9.8xx2xx1/2xx2/5`

`=v=sqrt(9.8xx1.6)=2.8*sqrt2m"/"s`

Answer 2

The speed is `=3.61ms^-1`

Explanation:

Another approach is to calculate the acceleration of the disc down the incline.

Let the coefficient if friction detween the disc and the incline be `=mu`

Then,

The acceleration down the plane is

`mgsintheta-mumgcostheta=ma`

`gsintheta-mugcostheta=a`......................`(1)`

Taking moments about the center of the disc

`tau=Ialpha`

`alpha=a/r`

`rmumgcostheta=1/2mr^2*a/r`

`mu=a/(2gcostheta)`............................`(2)`

Plugging this value in equation `(1)`

`gsintheta-a/(2gcostheta)gcostheta=a`#

Therefore,

`3/2a=gsintheta`

The acceleration is `a=2/3gsintheta`

The distance is `s=2m`

Apply the equation of motion

`v^2=u^2+2as`

`v^2=0+2as`

`v=sqrt(2*2/3gsintheta*s)`

`v=sqrt(4/3*9.8*sin30*2)=sqrt(4*9.8/3)=3.61ms^-1`

P.S There is a problem, the moment of inertia of the disc about the center is `I=1/2mr^2=1/2*2*1^2=1kgm^2` and not `0.5kgm^2`