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Standard XI

Physics

Direction of Friction

A small disc ...

Question

A small disc and a large disc are connected by a belt. The small disc whose radius is 15 cm rotates at an angular speed of 360 rpm. The radius of large disc having mass 8 kg is 45 cm. The kinetic energy of large disc is nearly equal to ( Assume $π^{2} = 10$ )

60 J

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64 J

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120 J

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45 J

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Solution

The correct option is B 64 J
Belt does not slip on wheels so, $v = r_{1} ω_{1} = r_{2} ω_{2} 15 (360) = 45 x x = 120 r p m = 2 r p s = 4 π r a d / s$

$K . E = \frac{1}{2} I ω^{2} = \frac{1}{2} \frac{m r^{2}}{2} ω^{2} = \frac{1}{2} (8) \frac{(0.45)^{2}}{2} (16 π^{2}) = 64 J$

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Similar questions

Q. A small disc and a large disc are connected by a belt. The small disc whose radius is 15 cm rotates at an angular speed of 360 rpm. The radius of large disc having mass 8 kg is 45 cm. The kinetic energy of large disc is nearly equal to ( Assume

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A small disc and a large disc are connected by a belt. The small disc whose radius is 15 cm rotates at an angular speed of 360 rpm. The radius of large disc having mass 8 kg is 45 cm. The kinetic energy of large disc is nearly equal to ( Assume π2=10)

The correct option is B 64 JBelt does not slip on wheels so, v=r1ω1=r2ω215(360)=45xx=120 rpm=2 rps=4π rad/s K.E=12Iω2=12mr22ω2=12(8)(0.45)22(16π2)=64J

A small disc and a large disc are connected by a belt. The small disc whose radius is 15 cm rotates at an angular speed of 360 rpm. The radius of large disc having mass 8 kg is 45 cm. The kinetic energy of large disc is nearly equal to ( Assume $π^{2} = 10$ )

The correct option is B 64 J
Belt does not slip on wheels so, $v = r_{1} ω_{1} = r_{2} ω_{2} 15 (360) = 45 x x = 120 r p m = 2 r p s = 4 π r a d / s$

$K . E = \frac{1}{2} I ω^{2} = \frac{1}{2} \frac{m r^{2}}{2} ω^{2} = \frac{1}{2} (8) \frac{(0.45)^{2}}{2} (16 π^{2}) = 64 J$