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This is the second of our many quizzes on Centripetal acceleration. Click on this to access more such competitive quizzes on Mechanics.

Centripetal acceleration Quiz 2

physics

A bob is attached to a string of radius $r$ and is spinning, the centripetal acceleration is given by, ${{a}_{c}}=\dfrac{{{v}^{2}}}{r}$. If the string is released, the motion of the bob is,

 Centripetal acceleration is the acceleration that acts on a body when it is traversing a circular path as shown in the figure. The acceleration is directed radially towards the center and the velocity is directed along the tangent on the circular path with velocity $v$. 

 Centripetal acceleration is the acceleration that acts on a body when it is traversing a circular path as shown in the figure. The acceleration is directed radially towards the center and the velocity is directed along the tangent on the circular path.If the string is released then the bob moves in a straight line along the tangent at that point on the circular path with a velocity $v$.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a1da221d-b221-40ad-913e-3f5a808e89145831851755015247979.png"/> 

Consider a centrifuge spinning at $24000$ rad/sec. What is the centripetal acceleration at a point that is $1$ m from the centre of the centrifuge? 

Since centripetal acceleration is proportional to square of the velocity and inversely proportional to the radius of the circular path, so to find it we use the formula which is given by ${{a}_{c}}=\dfrac{{{v}^{2}}}{r}=r{{\omega }^{2}}$.By substituting the radius and angular velocity of the centrifuge given in the question, we can find the centripetal acceleration at the point that is $1$ m from the centre of the centrifuge. 

 $\text{Centripetal acceleration is given by,}\\  {{a}_{c}}=\dfrac{{{v}^{2}}}{r}=r{{\omega }^{2}}….(i)\\\text{According to the question,}\\  \omega =24000\,rad/sec\,\text{and}\,r=1m\\\text{Putting the above values in equation (i), we get,}\\{{a}_{c}}=1\times {{(24000)}^{2}}\\  \therefore{{a}_{c}}=5.76\times {{10}^{6}}\,m/{{s}^{2}}$ 

 If the centripetal force acting on a yoyo of mass $45$ g spinning in a vertical circular path is $250$ N, then what is the centripetal acceleration of the bob? 

Relation between centripetal force and centripetal acceleration is given by $F=m{{a}_{c}}$.Find the centripetal acceleration using the values given in the question.

$\text{Relation between centripetal force and centripetal acceleration is given by,}\\ F=m{{a}_{c}}….(i)\\\text{According to the question,}\\ m=45\,g=0.045\,kg\,\text{and}\,F=250N\\\text{Putting the above values in equation (i), we get,}\\{{a}_{c}}=\dfrac{250}{0.045}\\  \therefore{{a}_{c}} =5.56\times {{10}^{3}}\,m/{{s}^{2}}$  

If a sphere of mass $35$ g tied to a string of length $15$ cm is moving in a vertical circular path, then the net maximum force exerted on the string is $220$ N. What is the centripetal acceleration and velocity of the sphere? 

 Centripetal force and centripetal acceleration are given by the equations $F=m{{a}_{c}}$ and ${{a}_{c}}=\dfrac{{{v}^{2}}}{r}$. We get the centripetal acceleration from the force formula and then to calculate the velocity of the sphere we put the value of centripetal acceleration and radius in the second equation and solve. 

$\text{Relation between centripetal force and centripetal acceleration is given by,}\\F=m{{a}_{c}}….(i)\\\text{According to the question,}\\  m=35\,g=0.035\,kg\,\text{and}\,F=220N\\  \text{Putting the above values in equation (i), we get,}\\  {{a}_{c}}=\dfrac{220}{0.035}\\\Rightarrow{{a}_{c}} =6.275\times {{10}^{3}}\,m/{{s}^{2}}\\  \text{Now, centripetal acceleration is given by,}\\  {{a}_{c}}=\dfrac{{{v}^{2}}}{r}\\\Rightarrow v=\sqrt{{{a}_{c}}\times r}….(ii)\\  \text{Putting the values of}\,{{a}_{c}}\,\text{and}\,r=15\,cm=0.15\,m\,\text{in equation (ii), we get,}\\  \therefore v=30.68\,m/s$   

Two bodies of mass ${{m}_{1}}=18$ g and ${{m}_{2}}=6$ g are attached together using a non stretchable wire. Mass ${{m}_{1}}$ is spinning with a radius $24$ cm and mass ${{m}_{2}}$ is suspended as shown in the figure. What should be the speed of mass ${{m}_{1}}$ so as the mass ${{m}_{2}}$ remains hanging? $(\text{ Take g}=9.8 m/{{s}^{2}})$.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ff8cd7f4-e6d4-40fe-8808-9d83de32d1af6054118408065542910.png"/>

Centripetal force is given by $F=m{{a}_{c}}=\dfrac{m{{v}^{2}}}{r}$. This force due to ${{m}_{1}}$ should be equal to the gravitational force on mass ${{m}_{2}}$, equate and calculate velocity of mass ${{m}_{1}}$. 

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c71f18c9-5229-4d0b-8cd8-429ca2fece437930855570601667773.png"/>$\\ \text{ As we see in the diagram, the balancing force that keeps the mass}\,{{m}_{2}}\,\text{hanging is the tension in the string.}\\  \text{This tension should be equal to the gravitational force on}\,{{m}_{2}}\,\text{as it remains hanging without accelerating.}\\\text{Centripetal force is given by,}\\ F=m{{a}_{c}}=\dfrac{m{{v}^{2}}}{r}….(i)\\ \text{ Since, the force F due to}\,{{m}_{1}}\,\text{ should be equal to the gravitational force on}\,{{m}_{2}},\\F=\dfrac{{{m}_{1}}{{v}^{2}}}{r}={{m}_{2}}g\\  \text{ So, velocity of}\,{{m}_{1}}\,\text{ will be,}\\  v=\sqrt{\dfrac{{{m}_{2}}gr}{{{m}_{1}}}}….(ii)\\  \text{According to the question,}\\{{m}_{1}}=18\,g=0.018\,kg,{{m}_{2}}=6\,g=0.006\,kg\,\text{and}\,r=24cm=0.24m\\ \text{ Putting the above values in equation (ii), we get,}\\ \therefore v=0.88\,m/s$