Torque

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A solid sphere can rotate about its diameter. When a torque is applied to this sphere about this diameter, it starts rotation with angular acceleration of $8 rad / s^{2}$ . What will be the angular acceleration when radius of the sphere is doubled and torque applied is also doubled ?

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Solution

$τ = Iα = (\frac{2}{5} {mr}^{2}) α = \frac{2}{5} . (\frac{4}{3} . {πr}^{3} ρ) r^{2} . α$
$\Rightarrow α = (\frac{15}{8 π ρ}) . \frac{τ}{r^{5}}$
$∴ \frac{α_{2}}{α_{1}} = (\frac{τ_{2}}{τ_{1}}) {(\frac{r_{1}}{r_{2}})}^{5} = (2) {(\frac{1}{2})}^{5} = \frac{1}{16}$
$\Rightarrow α_{2} = 8 x \frac{1}{16} rad / s^{2} = \frac{1}{2} rad / s^{2}$

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Torque

Remember concepts with our Masterclasses.

A solid sphere can rotate about its diameter. When a torque is applied to this sphere about this diameter, it starts rotation with angular acceleration of 8 rad/s2 . What will be the angular acceleration when radius of the sphere is doubled and torque applied is also doubled ?

τ=Iα=25mr2α=25.43.πr3ρr2.α⇒α=158πρ.τr5∴α2α1=τ2τ1r1r25=2125=116⇒α2=8 x 116rad/s2=12rad/s2

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A solid sphere can rotate about its diameter. When a torque is applied to this sphere about this diameter, it starts rotation with angular acceleration of $8 rad / s^{2}$ . What will be the angular acceleration when radius of the sphere is doubled and torque applied is also doubled ?