Question

If the two objects are released at rest, and the height of the ramp is h = 0.65 m, find the speed of the disk and the spherical shell when they reach the bottom of the ramp. Express your answers using two significant figures separated by a comma.

Answer 1

2.91 m/s, the speed of disk and 2.52 m/s, speed for hoop.

Step-by-step explanation:

The change in potential energy will equal the change in kinetic energy

`m * g * h=\left((1)/(2) * m * v^(2)\right)+\left((1)/(2) * I * w^(2)\right)`

With no slipping
`w=(v)/(r)`

The moment of inertia of a uniform disk is
`I=(1)/(2) * m * r^(2)`

`m * g * h=\left((1)/(2) * m * v^(2)\right)+\left((1)/(2) *\left((1)/(2) * m * r^(2)\right) *\left((v)/(r)\right)^(2)\right)`

`m * g * h=m\left((1)/(2) * v^(2)\right)+\left((1)/(4) * v^(2)\right)`

‘m’ gets cancelled on both sides, v squared taking as common on right side, we get,

`g * h=v^(2)\left((1)/(2)+(1)/(4)\right)=v^(2)\left((3)/(4)\right)`

we know,
`g=9.8 \mathrm{m} / \mathrm{sec}^(2)`

`v^(2)=(4)/(3) * 9.8 * 0.65 = 8.49`

Moment of inertia for a thin spherical shell (for hoop) is
`I=m r^(2)`

`m * g * h=\left((1)/(2) * m * v^(2)\right)+\left((1)/(2)\left(m * r^(2)\right)\left((v)/(r)\right)^(2)\right)`

Taking ‘m’ as common, and cancelling r squared, we get

`g * h=\left(v^(2)\right)\left((1)/(2)+(1)/(2)\right)`

`v^(2)=g * h=9.8 * 0.65=6.37`

Taking square root, we get

`v=√(6.37)=2.52 \mathrm{m} / \mathrm{s}`