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This is one of our many quizzes on Centripetal force. Click on this to access more such competitive quizzes on Mechanics.

Centripetal force Quiz 2

physics

Twirling a lasso is an example for _____ .

Centripetal force is the force acting when a body undergoes circular motion. 

 It is an example for the centripetal force. Here the centripetal force will be given by the tension over the rope which will pull the body in towards the centre. 

Find out the angular velocity of an object of mass $65\text{ }kg$ which is experiencing a centripetal force $65\text{ }N$ along a path of radius $90\text{ k}m$. 

The centripetal force will be the product of the mass, radius and the square of the angular velocity.  

 The equation of centripetal force can be written as, $ \\ {{F}_{C}}=mr{{\omega }^{2}} \\ $ Substituting the values in it will give, $ \\ 65=65\times 90\times {{10}^{3}}\times {{\omega }^{2}} \\  \Rightarrow {{\omega }^{2}}=\dfrac{1}{9\times {{10}^{4}}} \\ \Rightarrow \omega =\dfrac{1}{3}\times {{10}^{-2}}\text{ }rad{{s}^{-1}} \\ \therefore \omega =3.33\times {{10}^{-3}}\text{ }rad{{s}^{-1}}$ 

 Medha was enjoying a rollercoaster when she realised that it was going in a loop. When riding a rollercoaster, what will be the agent providing the centripetal force? Could you kindly assist her?

The normal force acting will give the centripetal force for the roller coaster. 

When a roller coaster is moving, the normal force as well as the gravity will be giving the necessary centripetal force to the motion through a loop. The downward movement will be because of the gravity.  

A centrifugal force is related to the inertia of a body. Identify whether the statement is correct or incorrect. 

Centrifugal force acts away from the circular path. 

Yes. The centrifugal force is having relation to the inertia of the body. Because of the inertia, the body will tend to have a straight line movement other than a rotation. Hence they are related. 

A particle of $5\text{ }kg$ is having a displacement vector which is given as $\vec{r}\left( t \right)=6\cos \omega t\hat{i}+5\sin \omega t\hat{j}$. The centripetal force acting will be _____ .

The acceleration will be the second derivative of the displacement with respect to time. 

The force can be found by the equation, $F=ma$ $ \\ $ As the displacement vector has been given, $ \\ \vec{r}\left( t \right)=6\cos \omega t\hat{i}+5\sin \omega t\hat{j} \\ $ The acceleration can be found from this, $ \\ \vec{r}\left( t \right)=6\cos \omega t\hat{i}+5\sin \omega t\hat{j} \\  \vec{v}\left( t \right)=\dfrac{d}{dt}\vec{r}\left( t \right) \\  \vec{v}\left( t \right)=-6\omega \sin \omega t\hat{i}+5\omega \cos \omega t\hat{j} \\ $ The acceleration will be the time derivative of the velocity, $\\ \vec{a}\left( t \right)=\dfrac{d}{dt}\vec{v}\left( t \right) \\  \Rightarrow \vec{a}\left( t \right)=-6{{\omega }^{2}}\cos \omega t\hat{i}-5{{\omega }^{2}}\sin \omega t\hat{j} \\   \Rightarrow \vec{a}\left( t \right)=-{{\omega }^{2}}\left( 6\cos \omega t\hat{i}+5\sin \omega t\hat{j} \right) \\  \therefore \vec{a}\left( t \right)=-{{\omega }^{2}}\vec{r}\left( t \right) \\ $ Substituting this in the equation, $ \\ F=m\left[ -{{\omega }^{2}}\vec{r}\left( t \right) \right] \\   \therefore F=-5{{\omega }^{2}}\vec{r}\left( t \right)$ 