# An object with a mass of  2 kg is traveling in a circular path of a radius of 4 m. If the object's angular velocity changes from  1 Hz to  6 Hz in  3 s, what torque was applied to the object?

Apr 9, 2017

88N

#### Explanation:

first we need to find out the change in instantaneous velocity .
which is from 1 Hz to 6 Hz using formula
$\omega$=$\frac{2 \pi R}{T} = v R$
$\implies v = 2 \frac{\pi}{T}$
substituting values of 1 Hz as $\frac{1}{\nu} = T$
we get the values for instantanious velocity for 2 cases of 1 Hz and
6 Hz as 6.28$m {s}^{- 1}$ and $39.26 m {s}^{- 1}$
we use the Newton's law of motion according to which
$v = u + a \cdot t$
substituting
u=6.28
v=39.26
and T=3
We get the value for 'a' Acceleration as 10.99$\cong$11$m {s}^{- 1}$
For torque =$m \cdot a \cdot R$
we substitute desired values and get 88 N