It could rotate while not advancing distance
Answer:
it have Potential energy
Explanation:
given data
Drag the pendulum to an angle 30∘
to find out
what form of energy does it have
solution
we know that pendulum start no kinetic energy when it release from any rest position then in starting it have potential energy only so that when pendulum is angle 30∘ at some height from ground so when it start it have potential energy same as in starting.
we know that the total energy is always conserve
so it have potential energy
Answer:
x=0.46m, speed=7.9m/s
Explanation:
Using the concept of conservation of energy:
1. kinetic energy of mass m and velocity v: 
2. gravitational potential energy of mass m, grav. acc. g and height h: 
3. potential energy in a spring with spring constant k and displacement from equilibrium x: 
Calculating x:


Calculating the speed:



Solving for
:

H2 or H:H
hope this helps
Answer:
Explanation:
Velocity of Top most point of wheel is twice the Velocity of centre of mass of wheel

Thus angular velocity is given by



![K.E.=\frac{1}{2}m_{stone}V^2+2\left [ \frac{1}{2}m_{cyl}\left [ \frac{V}{2}\right ]^2\right ]+2\left [ \frac{1}{2}I\omega ^2\right ]](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B1%7D%7B2%7Dm_%7Bstone%7DV%5E2%2B2%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7Dm_%7Bcyl%7D%5Cleft%20%5B%20%5Cfrac%7BV%7D%7B2%7D%5Cright%20%5D%5E2%5Cright%20%5D%2B2%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%20%5E2%5Cright%20%5D)
![K.E.=\frac{V^2}{2}\left [ m_{stone}+\frac{m_{roller}}{2}\right ]+I\omega ^2](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7BV%5E2%7D%7B2%7D%5Cleft%20%5B%20m_%7Bstone%7D%2B%5Cfrac%7Bm_%7Broller%7D%7D%7B2%7D%5Cright%20%5D%2BI%5Comega%20%5E2)
![K.E.=\frac{0.319^2}{2}\left [ 672+\frac{82}{2}\right ]+\frac{82\times 0.343^2\times 0.465^2}{2}](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B0.319%5E2%7D%7B2%7D%5Cleft%20%5B%20672%2B%5Cfrac%7B82%7D%7B2%7D%5Cright%20%5D%2B%5Cfrac%7B82%5Ctimes%200.343%5E2%5Ctimes%200.465%5E2%7D%7B2%7D)
