First slide

Rolling motion

Question

A wheel of mass 5 kg and radius 0.40 m is rolling on a road without sliding with angular velocity of 10 $rad s^{- 1}$ . The moment of inertia of the wheel about the axis of rotation is 0.65 kg $m^{2}$ . The percentage of kinetic energy of rotation in the total kinetic energy of the wheel is

Moderate

A

22.4 %
B

11.2 %
C

88.8 %
D

44.8%

Solution

Here, m = 5 kg, I = 0.65 kg $m^{2}$
$ω = 10 rad s^{- 1}, R = 0.40 m$
Linear velocity
v = R $ω$ = 0.40 $\times 10 = 4 {ms}^{- 1}$
Translatiional KE
= $\frac{1}{2} {mv}^{2} = \frac{1}{2} \times 5 \times 16 = 40 J$
Rotational KE
= $\frac{1}{2} {Iω}^{2} = \frac{1}{2} \times 0.65 \times 100 = 32.5 J$
Total KE = Translational KE + Rotational KE
= 40+32.5 = 72.5 J
Percentage of rotational KE = $\frac{Rotational KE}{Total KE} \times 100$
= $\frac{32.5}{72.5} \times 100 = 44.8 %$

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