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If a sphere is rolling, the ratio of tra...

If a sphere is rolling, the ratio of translational energy to total kinetic energy is given by:

Text Solution

Verified by Experts

The expression for total kinetic energy in pure rolling is,
`KE=KE_("TRANS")+KE_(NOT)`
For any object the total kinetic energy as per given equation
`KE=(1)/(2)MV_(CM)^(2)+(1)/(2)Mv_(CM)^(2)((K^(2))/(R^(2)))`
`KE=(1)/(2)Mv_(CM)^(2)(1+(K^(2))/(R^(2)))`
Then `(1)/(2)Mv_(CM)^(2)(1+(K^(2))/(R^(2)))=(1)/(2)Mv_(CM)^(2)+(1)/(2)Mv_(CM)^(2)((K^(2))/(R^(2)))`
The above equation suggests that in pure rolling the ratio of total kinetic energy, translational kinetic energy and rotational kinetic energy is given as,
`KE:KE_("TRANS"):KE_("ROT"): : (1+(K^(2))/(R^(2))):1: ((K^(2))/(R^(2)))`
Now, `KE_("TRANS"): KE_(ROT) : : 1: ((K^(2))/(R^(2)))`
For a solid spere, `(K^(2))/(R^(2))=(2)/(5)`
Then, `KE_("TRANS") : KE_("ROT") : : 1: (2)/(5) or KE_("TRANS") : KE_(NOT) : : 5:2`
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