Question

a ring rolls down an inclined plane without slipping .what is its velocity when it reaches the ground?

Answer 1

In this case you have:
Initial Kinetic Energy + Initial Potential Energy + Initial Rotational Kinetic Energy = Final Kinetic Energy + Final Potential Energy + Final Rotational Kinetic Energy.

Answer 2

Using the formulas for each and assuming no potential energy at the bottom and all potential energy at the top and released from rest you can solve the problem for V.

Answer 3

can u solve it 4 me

Answer 4

No that would be up for you to demonstrate your algebra skills and proper use of formulas.

Kinetic Energy:\[KE = \frac{ 1 }{ 2 }mv^2\]
Potential Energy:\[PE = mgh\]
Rotational Kinetic Energy:\[KE_R = \frac{ 1 }{ 2 } I \omega^2\]

Where,
m = mass
v = velocity
g = gravity
h = height
I = Inertia
Omega (w symbol) = angular velocity

Answer 5

\[\omega = \frac{ v }{ r }\]

Inertia of a ring: I am, yet again, assuming a thin ring:
\[I = mr^2\]
If you assume a thick hoop,
\[I = \frac{ 1 }{ 2 }mr^2\]

Answer 6

You now have enough information to solve for velocity using algebra and formulas.

Answer 7

thanks igot the answer it is v=\[\sqrt{4gh/3}\]