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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.

Respuesta :

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

          [tex]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)[/tex]

With no slipping [tex]w=\frac{v}{r}[/tex]

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

         [tex]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)[/tex]

         [tex]m \times g \times h=m\left(\frac{1}{2} \times v^{2}\right)+\left(\frac{1}{4} \times v^{2}\right)[/tex]

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

         [tex]g \times h=v^{2}\left(\frac{1}{2}+\frac{1}{4}\right)=v^{2}\left(\frac{3}{4}\right)[/tex]

we know, [tex]g=9.8 \mathrm{m} / \mathrm{sec}^{2}[/tex]

             [tex]v^{2}=\frac{4}{3} \times 9.8 \times 0.65 = 8.49[/tex]

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

           [tex]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)[/tex]

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

            [tex]g \times h=\left(v^{2}\right)\left(\frac{1}{2}+\frac{1}{2}\right)[/tex]

            [tex]v^{2}=g \times h=9.8 \times 0.65=6.37[/tex]

Taking square root, we get

            [tex]v=\sqrt{6.37}=2.52 \mathrm{m} / \mathrm{s}[/tex]

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