Question
Question: A string will break under a load of \(5\,Kg.\) to the one end of such a \(2\,m\) long string a mass ...
A string will break under a load of 5Kg. to the one end of such a 2m long string a mass of 1Kg is attached. The maximum rpm in the horizontal plane so that the string does not break is (g=10ms−2)
A. 28.66
B. 47.73
C. 38.92
D. 54.12
Solution
In order to find rpm of the string (rpm stands for rotation per minute), we will first find the magnitude of centripetal force acting on the load which is moving in circular path in horizontal plane and in order to continue this motion without breaking of string tis force will be balanced by the tension acting on string due to gravity.
Formula used:
Centripetal force on a body is calculated as
F=mrω2
where, m is the mass of the body, r is the radius of the circular path and ω is the angular velocity.
Frequency of the body is related with angular velocity as ω=2πν.
Complete step by step answer:
According to the question we have given that string will break under a load of 5Kg.
Tension on the string due to acceleration due to gravity is T=5×g
T=50N
Now, the mass of the load is m=1Kg. Length or radius of the circular path will be r=2m. And let ν be the frequency of load, using formula of centripetal force F=mrω2 and putting ω=2πν we get F=4π2ν2mr.
In order to balance the system F=T hence, on putting the values we will get,
4π2ν2mr=50
⇒ν2=7950
⇒ν=0.632
⇒ν=0.79s−1
Now, in order to find in rotations per minute we will multiply this value by 1min=60sec
So,
rpm=0.79×60
∴rpm=47.4min−1
And this value of rpm is very close to 47.73.
Hence, the correct option is B.
Note: It should be remembered that, the load will continue to move in circular motion as long as the centripetal force on load will be balanced by tension acting on string, otherwise the string will break and here, if we use the exact value of acceleration due to gravity g=9.8ms−2 we would get the exact value of rpm given in the option 47.73 but as in question its mentioned of using (g=10ms−2) that’s why our calculated value doesn’t exactly matched with the value given in option.