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.

Explanation:

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

          $m \times g \times h=\left(\frac{1}{2} \times m \times v^{2}\right)+\left(\frac{1}{2} \times I \times w^{2}\right)$

With no slipping $w=\frac{v}{r}$

The moment of inertia of a uniform disk is $I=\frac{1}{2} \times m \times r^{2}$

         $m \times g \times h=\left(\frac{1}{2} \times m \times v^{2}\right)+\left(\frac{1}{2} \times\left(\frac{1}{2} \times m \times r^{2}\right) \times\left(\frac{v}{r}\right)^{2}\right)$

         $m \times g \times h=m\left(\frac{1}{2} \times v^{2}\right)+\left(\frac{1}{4} \times v^{2}\right)$

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

         $g \times h=v^{2}\left(\frac{1}{2}+\frac{1}{4}\right)=v^{2}\left(\frac{3}{4}\right)$

we know, $g=9.8 \mathrm{m} / \mathrm{sec}^{2}$

             $v^{2}=\frac{4}{3} \times 9.8 \times 0.65 = 8.49$

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

           $m \times g \times h=\left(\frac{1}{2} \times m \times v^{2}\right)+\left(\frac{1}{2}\left(m \times r^{2}\right)\left(\frac{v}{r}\right)^{2}\right)$

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

            $g \times h=\left(v^{2}\right)\left(\frac{1}{2}+\frac{1}{2}\right)$

            $v^{2}=g \times h=9.8 \times 0.65=6.37$

Taking square root, we get

            $v=\sqrt{6.37}=2.52 \mathrm{m} / \mathrm{s}$