A string breaks under a load of 4.8 kg. A mass of 0.5 kg is attached t

1.	A string breaks under a load of 4.8 kg. A mass of 0.5 kg is attached to one end of the string 2 m longand is rotated in horizontal circle. Calculate the greatest number of revolutions that the mass can makewithout breaking the string.the string
Answer» Answer: The answer will be 66.12 RPM Explanation: According to the problem the load WEIGHT is 4.8 kg The mass is attached with the STRING having length of 2 m Now as the mass is rotating in horizontal circle, Then let the value of mass , m =0.5 kg and the radius of the circle is , r = 2m Let a tension T is acting in the center Therefore, T = mrω^2 [ where ω is the angular VELOCITY] Now we know ω = 2 πf Therefore, T = MR(2 πf)^2 Now from here if we calculate f f = √ T /mr x 4 π^2 Here T = 4.8 kg wt = 4.8 x 9.8 N f = √4.8 x 9.8 / 0.5 x 2 x 4 x 9.87 = 1.102 revolutions per second = 66.12 rpm

A string breaks under a load of 4.8 kg. A mass of 0.5 kg is attached to one end of the string 2 m longand is rotated in horizontal circle. Calculate the greatest number of revolutions that the mass can makewithout breaking the string.the string

Answer»

Answer:

The answer will be 66.12 RPM

Explanation:

According to the problem the load WEIGHT is 4.8 kg

The mass is attached with the STRING having length of 2 m

Now as the mass is rotating in horizontal circle,

Then let the value of mass , m =0.5 kg and the radius of the circle is , r = 2m

Let a tension T is acting in the center

Therefore,

T = mrω^2 [ where ω is the angular VELOCITY]

Now we know ω = 2 πf

Therefore,

T = MR(2 πf)^2

Now from here if we calculate f

f = √ T /mr x 4 π^2

Here T = 4.8 kg wt = 4.8 x 9.8 N

f = √4.8 x 9.8 / 0.5 x 2 x 4 x 9.87 = 1.102 revolutions per second = 66.12 rpm

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