Question

The maximum mass that can be hung verti from a string without breaking the string is TO KE. A length of this string that is 2 m long is used to rotate a 0.5 kg object in a circle on a frictionless table with the string horizontal. The maximum speed that the mass can attain under these conditions without the string breaking is most nearly (A) 5 m/s (B) 10 m/s (C) 14 m/s (D) 20 m/s (E) 100 m/s .: S

Answer 1

Explanation:

The maximum mass that can be hung from a string, m = 10 kg

Length of the string, l = 2 m

Mass of the object, m = 0.5 kg

Let v is the maximum speed that the mass can attain under these conditions without the string breaking. If T is the maximum tension in the string. So,

$T_{max}=mg$

$T_{max}=10\times 9.8=98\ N$

The centripetal force is provided by the tension in the string such that :

$T_{max}=\dfrac{mv^2}{r}$

$v=\sqrt{\dfrac{T_{max}r}{m}}$

$v=\sqrt{\dfrac{98\times 2}{0.5}}$

v = 19.79 m/s

or

v = 20 m/s

So, the maximum speed that the mass can attain under these conditions without the string breaking is 20 m/s. Hence, this is the required solution.