Question

A disc, a sphere and a ring with the same mass and the same radius roll down an inclined plane. A block also slides on an frictionless inclined plane of the same height. All objects start from the same height at rest. Rank the objects in terms of velocity of the center of mass at the bottom of the plane.

Answer 1

Block > Sphere > Disc > Ring

Step-by-step explanation:

Considering the cases for the rigid bodies disc, sphere and a ring in the problem, kinetic friction does no work if the bodies roll without slipping. We can assume that the bodies' surfaces and the surface of the incline are rigid and so ignore the effects of rolling friction. Each of the bodies start from the top of an incline with a height h, so KE1 = 0, PE1 = Mgh and PE2 = 0.

Since the bodies roll without slipping we have that ω = Vcm/R, the angular speed.

So Using the energy conservation equation for rigid bodies,

KE1 + PE1 = KE2 + PE2

0 + Mgh = (1/2×MVcm² + 1/2×I×ω²) + 0

Let I = cMR² = equal the moment of inertia for each of the bodies where c is a number equal to or less than 1.

Mgh = 1/2×MVcm² + 1/2×cMR²×(Vcm/R)²

Mgh = 1/2MVcm² + 1/2cMVcm²

Mgh = 1/2MVcm²(1 + c)

gh = 1/2Vcm²(1 + c)

Vcm² = 2gh/(1 + c)

Vcm = √(2gh/(1 + c))

So Case1: Disc

c = 1/2

Vcm = √(2gh/(1 + 1/2)) = √(2gh/(3/2))

Vcm = √(4/3×gh) = √(1.33×gh)

Case2: Sphere

c = 2/5

Vcm = √(2gh/(1 + 2/5)) = √(2gh/(7/5))

Vcm = √(10/7×gh) = √(1.43×gh)

Case2: Rings

c = 1

Vcm = √(2gh/(1 + 1)) = √(2gh/2)

Vcm = √(gh) = √(gh)

Now for the block of same mass M

The block starts from the top of the incline with a height h, so KE1 = 0, PE1 = Mgh and PE2 = 0. The incline is friction less so no work is done by kinetic frictional force.

KE1 + PE1 = KE2 + PE2

0 + Mgh = 1/2×MVcm² + 0

gh = 1/2×Vcm²

Vcm² = 2gh

Vcm = √(2gh)

So the velocities in order of greatest to least

Block > Sphere > Disc > Ring

Answer 2

(From the quickest to the slowest): Sphere, Disc, Ring.

Step-by-step explanation:

Let assume that height at the bottom is equal to zero. All elements can be described by means of the Principle of Energy Conservation:

`U = K_(t) + K_(r)`

`M\cdot g \cdot h = (1)/(2)\cdot M \cdot v^(2) + (1)/(2)\cdot I_(g) \cdot \left((v)/(R) \right)^(2)`

`g\cdot h = (1)/(2)\cdot \left[1 + (I_(g))/(M\cdot R^(2)) \right] \cdot v^(2)`

The velocity is determined by the following expression:

`v = \sqrt{(2\cdot g \cdot h)/(\left[1 + (I_(g))/(M\cdot R^(2)) \right]) }`

Case I - Disc (
`I_(g) = (1)/(2)\cdot M \cdot R^(2)`)

`v = \sqrt{(4\cdot g \cdot h)/(3) }`

`v\approx 1.155\cdot √(g\cdot h)`

Case II - Sphere (
`I_(g) = (2)/(5)\cdot M \cdot R^(2)`)

`v = \sqrt{(10\cdot g \cdot h)/(7) }`

`v\approx 1.195\cdot √(g\cdot h)`

Case III - Ring (
`I_(g) = M \cdot R^(2)`)

`v \approx √(g\cdot h)`