A solid sphere is released from height h from the top of an incline making an an

Question

A solid sphere is released from height h from the top of an incline making an angle ? with the horizontal.

(a) Calculate the speed of the sphere when it reaches the bottom of the incline in the case that it rolls without slipping. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
vf =  

(b) Calculate the speed of the sphere when it reaches the bottom of the incline in the case that it slides frictionlessly without rolling. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
vf =  


(c) Compare the time intervals required to reach the bottom in cases (a) and (b).
rolling time/sliding time =

Explanation / Answer

a)

for rolling,

let mass=m

height=h ,

angle of incline=theta

lenght of icline plane is l

then h=l*sin(theta)

now

by using conservation of energy

P.E=K.E+R.E

mgh=1/2*m*v^2+1/2*I*w^2

mgh=1/2*m*v^2+1/2*2/5*m*r^2*(v/r)^2

2gh=v^2+2/5*(v^2)

===>
V=sqrt[(10/7)*gh] .......
or
V=sqrt[(10/7)*gl*sin(theta)] .......

b)

for sliding

P.E=K.E

mgh=1/m*v^2

v=sqrt(2gh) .......

or

v=sqrt[2g*l*sin(theta)]

c)

in case of rolling

v^2=2*a*s

(10/7)*gh=2*a*l

===>

accelration a=(5/7)*gh/l

and

from

s=(1/2)*at^2

t=sqrt(2l/a)

=sqrt(2l/(5/7)*gh/l)

=sqrt[14*l^2/5g*l*sin(theta)]

tr=sqrt[(14l/5g*sin(theta)] .......(in case of rolling)

now

in case of sliding

v^2=2*a*s

2gh=2*a*l

====>
acceleration a=g*h/l

=g*l*sin(theta)/l

=g*sin(theta)

from

s=(1/2)*at^2

t=sqrt(2l/a)

ts=sqrt(2l/gsin(theta) ........(in case of slipping)

tr/ts=sqrt[14l/5g*sin(theta)]/sqrt[2l/gsin(theta)]

tr/ts=sqrt(7/5)

tr/ts=1.18

tr=1.18*ts

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