Question

a disc at rest rolls down a hill of height (3×9.8) without slipping .what is its velocity when it reaches the bottom

Answer 1

Answer:

The easiest way to solve these problems is to apply conservation of energy. The disc at rest only has potential energy. If we define the datum (h=0) to be the bottom of the hill, then the disc will have no potential energy at the bottom of the hill. All the potential energy will be converted to linear kinetic energy as well as rotational kinetic energy.

PE=(KE)linear+(KE)rotational

mgh=12mv2+12Iω2 ——- equation 1

but the mass moment of inertia of a disc about its rotating axis is I=12mR2

and the angular velocity ω=vR

equation 1 becomes:

mgh=12mv2+12(12mR2)(vR)2

or

mgh=12mv2+14mv2

or

mgh=34mv2

or

gh=34v2

(9.8)(3)9.8=34v2

or

(9.8)9.8=14v2

take square root of both sides gives:

9.8=12v

v=2(9.8)=19.6m/s