Question

we have got two vertical track of equal radius . one is smooth and other is rough .one small solid sphere is to be projected from the bottom of the circular tracks.it is seen that when it is projected on the smooth track the minimum linear speed is u1 to reach the top of the smooth track and in case rough track it is u2 .Ratio of u1 to that of u2 is​

Answer 1

Answer:

ANSWER

K.E=P.E

2

1

​  

mv  

2

[1+  

r  

2

 

k  

2

 

​  

]=mgh

v  

2

[1+  

3

2

​  

]=gh

v=  

5

3

​  

gh

​  

 

Now  

2

1

​  

mv  

2

+RotationalK.E=mgh

∴mgh−  

2

1

​  

m×  

5

3

​  

gh= Rotational K.E

So Rotational Kinetic Energy  

=(1−  

10

3

​  

)mgh

=  

10

7

​  

mgh

Explanation:

see in specific method