Question

a solid sphere of mass M and radius R rolls on a rough horizontal plane speed u, now it rolls on a smooth inclined plane of inclination 30 degree. the length of inclined plane travelled by sphere

Answer 1

Answer:

_________________________

T.E.=Rot.K.E.+Trans.K.E.

Rot.K.E.=(1/2)Iw²=(1/2)(2/5)Mr²w²=(1/5)Mv²

Tans.K.E.=(1/2)Mv²

So Total energy=(1/2)Mv²+(1/5)Mv²=(7/10)Mv².

So the % of R.K.E. is(1/5)Mv²/{(7/10)Mv²}×100

=(2/7)×100=28.57%.Ans

Answer 2

Answer:

The length of inclined plane traveled by sphere is 3.83 m

Explanation:

Speed of center of mass (v)= 5m/s  

Angle of Inclination ө= 30°  

a) The cylinder that is rolling up the inclined plane has an acceleration given by  

$\mathrm{a}=\frac{-g \sin \theta}{\left(1+k^{2} / r^{2}\right)}$

K = gyration radius and MI of the cylinder $= \frac{1}{2} m \times r^{2}$

Then $\mathrm{K}^{2}=\mathrm{r}^{2} / 2$

$\begin{array}{l}{\mathrm{a}=\frac{-g \sin 30^{\circ}}{\left(1+\frac{1}{2}\right)}} \\ {=-\frac{9.8}{3} \mathrm{m} / \mathrm{s}^{2}}\end{array}$

Applying equation of motion in the next step,  

$\begin{array}{l}{\mathrm{V}^{2}=\mathrm{U}^{2}+2 \mathrm{aS}} \\ {0=(5)^{2}+2 \mathrm{x}\left(-\frac{9.8}{3}\right) \mathrm{x} \mathrm{S}}\end{array}$

S = 3.83 m  

Let it return in time “t”

$S=u \times t+\frac{1}{2} \times a \times t^{2}$

t = 1.53 sec