I have this question, in my physics book where a bob attached to a string is in a slanting position as it is inside a cart moving forward with an acceleration. I understand that if we draw the free body diagram of the bob from inside the cart, we have to include a pseudo force as we are applying newtons laws from a non-inertial frame of reference. My question is, can we draw the same free body diagram from an inertial frame of reference, i.e. the road? if we can draw it, can someone tell me what the forces acting on the bob are? are we are supposed to get the same free body diagram as the previous case? THis is confusing me for some reason!
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Is it a sort of pendulum? You will only get the weight and the tension then. $\endgroup$– MauricioCommented Nov 21, 2017 at 8:10
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
From the point of view of somebody in the non-inertial frame, the bob is stationary. Accordingly, the net force on the bob (tension + gravity + pseudo-force) is equal to zero.
From the point of view of an inertial observer, there is no pseudo-force, so the net force on the bob (tension + gravity) is not equal to zero. This makes sense, because the bob is indeed accelerating in such a coordinate system.
answered Nov 21, 2017 at 8:12
-
$\begingroup$ Let us take the angle made by the string with the vertical as b. Let a be the acceleration of the cart and m be the mass of the bob. So for the non-inertial observer, the equation along x axis would be Tsin(b)=ma. ma is the pseudo force acting in a direction opposite to the motion of the cart and Tsinb, component of tension along x axis. Now, from the intertial reference, how will explain the slanted position of the bob without adding pseudo force? you might tell me that since the cart is moving, the bob is slanted because of inertia, so with the parameters i have given, can you writeanequatio $\endgroup$ Commented Nov 21, 2017 at 8:32
-
$\begingroup$ @GouthamSwaminathan $\sum F_x = T\sin(\theta) = ma_x$, so $a_x = \frac{T\sin(\theta)}{m}$. I'm not sure what else you're looking for. $\endgroup$ Commented Nov 21, 2017 at 8:51
-
$\begingroup$ @GouthamSwaminathan In the non inertial frame you have the equation $T\sin b -ma_x =0$ while in the inertial frame you have the equation $T\sin b = ma_x$, where the lhs is the forces acting on the bob in each respective frame. So the equations are of course mathematically equivalent, it’s just their physical interpretation that’s different. $\endgroup$– CAFCommented Nov 21, 2017 at 9:26