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The Earth is revolving around sun with an angular velocity w in an orbit of radius D and rotating around its own axis with an angular velocity w' . What ios its total kinetic energy? As earth is a sphere and different particles are at different distances from the axis of the sun , how we can consider the c.m velocity of the earth ? y cant we apply parallel axis theorem to get the answer as 0.5 (0.4MR2+MD2)w2+0.5 (0.4MR2)w'2
The Earth is revolving around sun with an angular velocity w in an orbit of radius D and rotating around its own axis with an angular velocity w' . What ios its total kinetic energy?
As earth is a sphere and different particles are at different distances from the axis of the sun , how we can consider the c.m velocity of the earth ? y cant we apply parallel axis theorem to get the answer as 0.5 (0.4MR2+MD2)w2+0.5 (0.4MR2)w'2
When you apply parallel axis theorem (and get the answer as 0.5 (0.4MR2+MD2)w2+0.5 (0.4MR2)w'2), you place the earth at sun's position with angular velocity w and then you displace the earth to its orbit and apply the parallel axis theorem to get the answer, but you don't care about the angular velocity. In original question the earth is already in its orbit and rotating with w. To calculate the kinetic energy, apply K.E. = K.E. of earth with respect to c.m. + K.E. of c.m. with respect to the sun Similar to the case when you calculate abgular momentum, viz, L = Lcm + rXmvcm The answer to the question will be 0.5 (0.4MR2)w'2 +0.5(MD2)w2
When you apply parallel axis theorem (and get the answer as 0.5 (0.4MR2+MD2)w2+0.5 (0.4MR2)w'2),
you place the earth at sun's position with angular velocity w and then you displace the earth to its orbit
and apply the parallel axis theorem to get the answer, but you don't care about the angular velocity.
In original question the earth is already in its orbit and rotating with w.
To calculate the kinetic energy, apply
K.E. = K.E. of earth with respect to c.m. + K.E. of c.m. with respect to the sun
Similar to the case when you calculate abgular momentum, viz,
L = Lcm + rXmvcm
The answer to the question will be 0.5 (0.4MR2)w'2 +0.5(MD2)w2
IF IT IS THE CASE OF A THIN UNIFORM ROD OF MASS M , LENGTH L ROTATING AROUND SUN SO THAT ITS ONE END IS AT R-(L/2) AND ANOTHER END IS AT R + (L/2) WITH AN ANGULAR VELOCITY ω . WHAT IS ITS KINETIC ENERGY? I GOT THE ENERGY AS 0.5 ((ML2 /12)+ MR2 )ω2 IS IT CORRECT? KINDLY GIVE ME THE ANSWER
IF IT IS THE CASE OF A THIN UNIFORM ROD OF MASS M , LENGTH L ROTATING AROUND SUN SO THAT ITS ONE END IS AT R-(L/2) AND ANOTHER END IS AT R + (L/2) WITH AN ANGULAR VELOCITY ω . WHAT IS ITS KINETIC ENERGY?
I GOT THE ENERGY AS 0.5 ((ML2 /12)+ MR2 )ω2 IS IT CORRECT? KINDLY GIVE ME THE ANSWER
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