The coin placed on a revolving disc,with its centre at distance of 6cm from the axis of rotation just slips off when the speed of the be the maximum angular speed of the revolving disc exceeds 45rpm .What should be the maximum angular speed of the disc ,so that when the coin is at a distance of 12cm from the axis of rotation , it does not slip?
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Given: = 15 cm = 0.15 m
60
nj = 60 r.p.m = = = 60 I r.p.s.
o, = 2 n n, = 2 n rad/s =
= 2 x 3.14 =
= 6.28 rad/s
75
m = 75 r.p.m. = r.p.s
60
= 1.25 r.p.s
(0) = 1.25 x 2 x 3.14
= 7.85 rad/sec
To find: Distance of the coin from the axis
(r2)
Formula: mr, - mrp
Calculation:
From formula
0.15x(6.28)
(7.85)
= antilog [log (0.15) + log(6.28) - log(7.85sy] = antilog (log(0.15) +2 log(6.28)
- 2 log (7.85)
= antilog [ T.1761+2 × 0,7980 - 2 x 0.8949]
- antilog [T.1761
= antilog (0.7726 +2.2102] –
+ 1.5960
= antilog (2.9828]
= 0,09612 m = 0.096 m
T- 9.6 cm
The distance of the coin from the axis so that it does not slip-off is 9.6 cm.
1.7898]
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