Question

15. A disc at rest without slipping, rolls down a hill of height (3 x 9.8) m. What is its speed in m/sec
when it reaches at the bottom?
A) 11.4
C) 22.8
B) 19.6
D) 9.8
​

Answer 1

Answer:

d) 9.8

Explanation:

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)rotationalPE=(KE)linear+(KE)rotational

mgh=12mv2+12Iω2mgh=12mv2+12Iω2 ——- equation 1

but the mass moment of inertia of a disc about its rotating axis is I=12mR2I=12mR2

and the angular velocity ω=vRω=vR

equation 1 becomes:

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

or

mgh=12mv2+14mv2mgh=12mv2+14mv2

or

mgh=34mv2mgh=34mv2

or

gh=34v2gh=34v2

(9.8)(3)9.8=34

Answer 2

Answer:

19.6 m/s

Explanation:

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=1/2mv2+1/2Iω2 ——- equation 1

but the mass moment of inertia of a disc about its rotating axis is I=1/2mR2

and the angular velocity ω=v/R

equation 1 becomes:

mgh=1/2mv2+1/2(1/2mR2)(v/R)2

or

mgh=1/2mv2+1/4mv2

or

mgh=3/4mv2

or

gh=3/4v2

(9.8)(3)9.8=3/4v2

or

(9.8)9.8=1/4v2

take square root of both sides gives:

9.8=1/2v

v=2(9.8)=19.6m/s