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

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

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. 1470 J
B. 1452 J
C. 1567J
D. 1632J

Monjur Alahi

5 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 - A
Work done `= Delta K.E. = (1)/(2)I(omega_(2)^(2)-omega_(1)^(2))`
`= (9.8)/(2xx pi^(2))xx 4pi^(2)`
`(n_(1)^(2)-n_(2)^(2))=19.6xx75`
= 1470 J.

5 views Answered 2023-02-05 15:29