Answer
Verified
97.5k+ views
Hint: The equations formed by the free body diagram can be used to make equations and by solving these equations we can solve this problem. The constant acceleration of the car will make the suspended ball to make an angle $\theta $ with the vertical.
Complete step by step solution:
(i) It is given in the problem that a bob of mass m is suspended in the car which is accelerating with the acceleration $\alpha $ and we need to find the angle of the bob with vertical.
From the above free body diagram.
In the vertical direction,
$ \Rightarrow T\cos \theta = mg$
$ \Rightarrow T = \dfrac{{mg}}{{\cos \theta }}$………eq. (1)
Also,
In the horizontal direction,
$ \Rightarrow T\sin \theta = ma$
$ \Rightarrow T = \dfrac{{ma}}{{\sin \theta }}$………eq. (2)
Equating the tension from equation (1) and equation (2) we get.
$ \Rightarrow \dfrac{{mg}}{{\cos \theta }} = \dfrac{{ma}}{{\sin \theta }}$
$ \Rightarrow \dfrac{g}{{\cos \theta }} = \dfrac{a}{{\sin \theta }}$
$ \Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{a}{g}$
$ \Rightarrow \tan \theta = \dfrac{a}{g}$
$ \Rightarrow \theta = {\tan ^{ - 1}}\dfrac{a}{g}$.
(ii) The angle of the bob with the string which is inside the car makes and angle $\theta $ with the vertical is equal to, $\theta = {\tan ^{ - 1}}\dfrac{a}{g}$.
Now let the angle of the inclination of the inclined plane be $\phi $.
From the above free body diagram,
In the direction perpendicular to the inclined plane.
$ \Rightarrow ma\cos \phi = mg\sin \phi $
$ \Rightarrow a\cos \phi = g\sin \phi $
$ \Rightarrow \dfrac{{\sin \phi }}{{\cos \phi }} = \dfrac{a}{g}$
$ \Rightarrow \tan \phi = \dfrac{a}{g}$
$ \Rightarrow \phi = {\tan ^{ - 1}}\dfrac{a}{g}$
The angle of inclination is given by$\phi = {\tan ^{ - 1}}\dfrac{a}{g}$.
The angle of the bob with vertical which is inside the car is equal to $\theta = {\tan ^{ - 1}}\dfrac{a}{g}$ and the angle of inclination of the smooth inclined plane is equal to $\phi = {\tan ^{ - 1}}\dfrac{a}{g}$.
Note: The bob experiences the forces on the backwards as the response of the acceleration of the car and at the equilibrium position it holds making an angle $\theta $ .The tension in the string always acts away from the body. The acceleration of the car is in the horizontal direction and the weight of the car is in the vertically downward direction.
Complete step by step solution:
(i) It is given in the problem that a bob of mass m is suspended in the car which is accelerating with the acceleration $\alpha $ and we need to find the angle of the bob with vertical.
From the above free body diagram.
In the vertical direction,
$ \Rightarrow T\cos \theta = mg$
$ \Rightarrow T = \dfrac{{mg}}{{\cos \theta }}$………eq. (1)
Also,
In the horizontal direction,
$ \Rightarrow T\sin \theta = ma$
$ \Rightarrow T = \dfrac{{ma}}{{\sin \theta }}$………eq. (2)
Equating the tension from equation (1) and equation (2) we get.
$ \Rightarrow \dfrac{{mg}}{{\cos \theta }} = \dfrac{{ma}}{{\sin \theta }}$
$ \Rightarrow \dfrac{g}{{\cos \theta }} = \dfrac{a}{{\sin \theta }}$
$ \Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{a}{g}$
$ \Rightarrow \tan \theta = \dfrac{a}{g}$
$ \Rightarrow \theta = {\tan ^{ - 1}}\dfrac{a}{g}$.
(ii) The angle of the bob with the string which is inside the car makes and angle $\theta $ with the vertical is equal to, $\theta = {\tan ^{ - 1}}\dfrac{a}{g}$.
Now let the angle of the inclination of the inclined plane be $\phi $.
From the above free body diagram,
In the direction perpendicular to the inclined plane.
$ \Rightarrow ma\cos \phi = mg\sin \phi $
$ \Rightarrow a\cos \phi = g\sin \phi $
$ \Rightarrow \dfrac{{\sin \phi }}{{\cos \phi }} = \dfrac{a}{g}$
$ \Rightarrow \tan \phi = \dfrac{a}{g}$
$ \Rightarrow \phi = {\tan ^{ - 1}}\dfrac{a}{g}$
The angle of inclination is given by$\phi = {\tan ^{ - 1}}\dfrac{a}{g}$.
The angle of the bob with vertical which is inside the car is equal to $\theta = {\tan ^{ - 1}}\dfrac{a}{g}$ and the angle of inclination of the smooth inclined plane is equal to $\phi = {\tan ^{ - 1}}\dfrac{a}{g}$.
Note: The bob experiences the forces on the backwards as the response of the acceleration of the car and at the equilibrium position it holds making an angle $\theta $ .The tension in the string always acts away from the body. The acceleration of the car is in the horizontal direction and the weight of the car is in the vertically downward direction.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
An object is moving with speed v0 towards a spherical class 12 physics JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
The capacity of a pure capacitor is 1 farad In dc circuits class 11 physics JEE_Main