Circular Motion
10.0 Centrifugal force
10.0 Centrifugal force
It is a pseudo force, which comes into play when we observe a circular motion from a rotating or non-inertial frame of reference.
Let us understand it briefly by comparing it with the centripetal force.
| Centripetal force | Centrifugal force |
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| In inertial frame of reference Newton laws of motion can be applied. | In non-inertial frame of reference Newton laws of motion cannot be applied. |
| In this case, the observer is at inertial frame of reference. Let the observer be at rest. Velocity of observer, $v_o=0$ Velocity of block, $v_b=v$ Velocity of block w.r.t. observer, $v_{bo}=v$ | In this case, the observer is at non-inertial frame of reference. Let the observer is sitting on block. Velocity of observer, $v_o=v$ Velocity of block, $v_b=v$ Velocity of block w.r.t. observer, $v_{bo}=0$ |
| Centripetal force as observed by observer, $${a_c} = \frac{{m{v^2}}}{r}$$ | Centripetal force as observed by observer, $${a_c} = 0$$ As the velocity of block w.r.t. observer is zero |
| From FBD, $$T = \frac{{m{v^2}}}{r}$$ | From FBD : (Incorrect observation) $T$ is unbalanced, which is incorrect So, a pseudo force known as centrifugal force is applied. Centrifugal force is applied in the direction apposite to the centripetal force. Centrifugal force converts the non-inertial frame of reference into inertial frame of reference. So, the Newton laws of motion can be applied. From FBD : Corrected observation $$T = \frac{{m{v^2}}}{r}$$ Here, $\left( {\frac{{m{v^2}}}{r}} \right)$ is the centrifugal force |
| The direction of centripetal force is towards the center $O$ and has a magnitude of $\left( {\frac{{m{v^2}}}{r}} \right)$ | The direction of centrifugal force is opposite to the center $O$ and has a magnitude of $\left( {\frac{{m{v^2}}}{r}} \right)$ |
Question 11. A system is rotating with an angular velocity of $\omega $ as shown in the figure. Find the minimum angular velocity $\omega $ and tension in the string when the blocks are at the verge of sliding.
The coefficient of friction between the blocks is $\mu $ and all other surfaces are smooth.
Solution: As the system is rotating with angular velocity $\omega $.
Let an observer is sitting at the top of the tower, which is rotating with same angular velocity.
Since the observer is rotating, it is a non-inertial frame of reference.
So, from FBD we get, $$\begin{equation} \begin{aligned} T + f = {m_1}{\omega ^2}L\quad ...(i) \\ {N_1} = {m_1}g\quad ...(ii) \\ T = f + {m_2}{\omega ^2}L\quad ...(iii) \\ {N_2} = {m_2}g + {N_1}\quad ...(iv) \\\end{aligned} \end{equation} $$
When the system is at the verge of sliding, limiting friction acts between the blocks.
Therefore, $$\begin{equation} \begin{aligned} f = {f_L} \\ f = \mu {N_1}\quad ...(v) \\\end{aligned} \end{equation} $$
From equation $(ii)$ and $(v)$ we get,
$$f = \mu {m_1}g\quad ...(vi)$$
From equation $(i)$, $(iii)$ and $(vi)$ we get,
$$\omega = \sqrt {\frac{{2\mu {m_1}g}}{{\left( {{m_1} - {m_2}} \right)L}}} \quad ...(vii)$$
From equation $(iii)$, $(v)$ and $(vii)$ we get, $$\begin{equation} \begin{aligned} T = \mu {m_1}g + \frac{{2\mu {m_1}{m_2}g}}{{\left( {{m_1} - {m_2}} \right)}} \\ T = \frac{{\mu {m_1}\left( {{m_1} + {m_2}} \right)g}}{{\left( {{m_1} - {m_2}} \right)}} \\\end{aligned} \end{equation} $$
