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प्रश्न
A motorcyclist (as a particle) is undergoing vertical circles inside a sphere of death. The speed of the motorcycle varies between 6 m/s and 10 m/s. Calculate the diameter of the sphere of death. How much minimum values are possible for these two speeds?
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उत्तर
Given:
`v_"top"` = 6 m/s,
`v_"bot"` = 10 m/s,
g = 10 m/s2
To find:
- the diameter of the sphere
- minimum values are possible for the two speeds
Solution:
`v_"bot"^2 = v_"top"^2 + 4gr`
∴ r = `(v_"bot"^2 - v_"top"^2)/(4g)`
`=((10)^2 - (6)^2)/(4 xx 10)`
= `64/40`
r = 1.6 m
Diameter = 2r
∴ The diameter of the sphere of death = 3.2 m
(ii) `v_"min" = sqrt(gr)` at the top.
∴ `v_"min" = sqrt(10 xx 1.6)`
= `sqrt16`
= 4 m/s
The corresponding minimum speed at the bottom
= `sqrt (5gr)`
= `sqrt (5(10)(1.6))`
= `sqrt80`
= `4 sqrt 5` m/s
The required minimum values of the speeds are 4 m/s and `4 sqrt 5` m/s.
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