Question

A solid disk is rolling without slipping on a level surface at a constant speed of 2.50 m/s. (a) if the disk rolls up a 30.0o ramp, how far along the ramp will it move before it stops? (b) explain why your answer in part (a) does not depend on either the mass or the radius of the disk.

Answer 1

(a) By using the law of conservation of energy, the distance moved by the solid disk along the ramp is 0.96 m.

(b) The answer does not depend on the mass and radius of the disc as these values are canceled when the law of conservation of energy is applied.

What is the law of conservation of energy?

The law of conservation of energy states that the total energy of an isolated system is conserved.

The total initial energy Ei of the disc rolling down an inclined plane is given by the formula,

Ei=1/2*mv^2 + 1/2*Iω^2

where m is the mass of the disc, v is the velocity of the disc, I is the moment of inertia and ω is the angular velocity of the disc.

For a solid disc, I=1/2mr^2, and since it is rolling without slipping, the rolling velocity of the disc will be equal to its translational velocity, that is,

v=ωr or ω=v/r

So using ω=v/r and I=1/2mr^2, it can be written,

Ei=1/2*m(v)^2 + 1/2*(1/2mr^2)(v/r)^2

Ei=1/2*m(v)^2*(1+1/2*)

Ei=3/4*mv^2

If the height covered by the disc is h before stopping, then its final total  energy Ef will be equal to the potential energy, that is,

Ef=m*g*h

From the law of conservation of energy, it can be written,

Ei=Ef

3/4*mv^2=m*g*h

h=3v^2/(4g)

The length of the ramp is then given by the formula,

l=hcosecθ

where θ is the inclination angle. So

l= 3v^2/(4g)*cosecθ

Here g=9.8 m/s^2, v=2.5 m/s  and θ=30.0 degree. Using these values,

l= 3*(2.5 )^2/(4*9.8)*cosec( 30)

l= 0.96 m

Learn more about the law of conservation of energy,

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