coefficient of static friction between a coin and a gramophone disc is 0.5 radius of the disc is 8cm initially the centre of the coin is pie cm away from the centre of the disc at what minimum frequency will it start slipping from there by what factor will the answer change if the coin is almost at the rim . use (g is equql to pie square)
Answers
Explanation:
Coefficient of static friction between a coin and a gramophone disc is .5 .
Radius of disc = 8 cm.
Center of coin is 2 cm away from the center of disc.
To Find:
At what minimum frequency will the coin start slipping from the gramophone disc.
Solution:
When a body is rotating about a axis , it experiences a centrifugal force which is equal to mw^{2} rmw
2
r , where
m is the mass of body , w is the frequency of rotation of body about the axis and r is the radial distance or the radius at which the body rotates about the axis.
The coin is at equilibrium or at a verge of motion .
Hence net forces in horizontal and vertical direction is 0.
vertical direction :
N = mg
horizontal direction :
friction force between disc and coin = centrifugal force acting on the coin (So that is in on the verge of motion )
friction force = \mu NμN
where \muμ is static friction coefficient between disc and coin. = 0.5
centrifugal force = mw^{2} rmw
2
r
r = 2 cm
\begin{lgathered}\mu N = m w^{2}r\\ 0.5(mg)= m(2\pi f )^{2} r\\0.5(\pi ^{2} )= m (4\pi ^{2}f^{2} (.02)\\ f = 2.5 cycles per seconds.\end{lgathered}
μN=mw
2
r
0.5(mg)=m(2πf)
2
r
0.5(π
2
)=m(4π
2
f
2
(.02)
f=2.5cyclesperseconds.
b) If coin is at rim
r = 8 cm.
\}5(mg) = 4m \pi ^{2} f^{2} r\\r = .08 m\\f = 1.25 cycles per second
.5(mg)=4mπ
2
f
2
r
r=.08m
f=1.25cyclespersecond
a ) f = 2.5 cycles per second . b) f = 1 .25 cycles per second.
plz mark it as brainlliest answer
Answer:
Aloha !
ūN= mw²r
Where N= mg
0.5×m.g= m(2πf)²r
0.5×π²= m×4π²f²r
0.5×π²= m×4π²f²×0.2
f=2.5
Hope it will help you
@ Sweety Adihya
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