For circular motion in the vertical plane, what is the centripital acceleration at the bottom of the circle?
What is the magnitude of the centripetal acceleration at the top of the trajectory ?
express your ans. in terms of g, M and T,
where is T is tension and M is mass of the object. If we move the mass attached to a rope in vertical circular motion
centripetal acceleration is v^2/r everywhere on the circle.
Tension=ma+-mv^2/r
at the bottom, +, at the top, -
To find the centripetal acceleration at the bottom of the circle in vertical circular motion, we need to consider the forces acting on the object. In this case, there are two forces: the tension force T in the rope and the gravitational force mg.
At the bottom of the circle, the tension force and the gravitational force act in the same direction, both vertically downward. Therefore, the net force is the sum of these two forces: F = T + mg.
The centripetal acceleration at the bottom of the circle is given by the equation a = F/m. Substituting the expression for the net force, we have:
a_bottom = (T + mg)/m
Now let's find the magnitude of the centripetal acceleration at the top of the trajectory. At the top, the tension force and the gravitational force act in opposite directions. The net force is the difference between these two forces: F = T - mg.
Using the same equation for centripetal acceleration, we get:
a_top = (T - mg)/m
These equations express the centripetal acceleration at the bottom and the top of the circle in terms of T, M, and g.