A disc of moment of inertia `9.8//pi^(2)kg m^(2)` is rotating at 600 rpm. If the frequency of rotation changes from 600 rpm to 300 rpm, then what is t

Question

A disc of moment of inertia `9.8//pi^(2)kg m^(2)` is rotating at 600 rpm. If the frequency of rotation changes from 600 rpm to 300 rpm, then what is t

A disc of moment of inertia `9.8//pi^(2)kg m^(2)` is rotating at 600 rpm. If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done ?
A. 1567 J
B. 1452 J
C. 1467 J
D. 1632 J

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
`omega = KE_(2)-KE_(1)=(1)/(2)I (omega_(2)^(2)-omega_(1)^(2))`
`= (1)/(2)xx(9.8)/(pi^(2))xx4 pi^(2)(n_(2)^(2)-n_(1)^(2))`
`= 4.9xx4(25-100)`
`=4.9xx4xx75`
`= 4.9xx300`
`= 1470.0`
`= 1467 J`

6 views Answered 2023-02-05 15:29