Centripetal Acceleration Question

AI Thread Summary
An object weighing 4 Newtons swings as a pendulum with a tension of 6 Newtons at the bottom of the swing. The centripetal acceleration can be derived from the equation Sum of forces = T - mg = mAc, where T is tension and mg is the weight. The correct approach involves recognizing that centripetal acceleration is not a force but an acceleration, which should be placed on the right side of the equation. To find the centripetal acceleration, the mass must first be calculated from the weight. The discussion emphasizes the importance of correctly applying Newton's second law and understanding the distinction between forces and acceleration.
tdreceiver17
Messages
10
Reaction score
0
An object weighing 4 Newtons swings on the end of a string as a simple pendulum. At the bottom the swing, the tension in the string is 6 Newtons. What is the magnitude of the centripetal acceleration of the object at the bottom of the swing.Centripetal Acc. = v^2/r Sum of forces = T+Ac=mg?Attempt

-T-mg=Ac?
-6-4 = Ac
I don't know where to go from here.
 
Physics news on Phys.org
Newton's second law says
$$\sum \vec{F}_i = m\vec{a}.$$ The centripetal acceleration is not a force. It's an acceleration. It goes into the righthand side of F=ma. The tension and the weight go into the lefthand side. Try again and pay attention to the sign of T and mg when summing the forces.
 
vela said:
Newton's second law says
$$\sum \vec{F}_i = m\vec{a}.$$ The centripetal acceleration is not a force. It's an acceleration. It goes into the righthand side of F=ma. The tension and the weight go into the lefthand side. Try again and pay attention to the sign of T and mg when summing the forces.

So I set it up as Sum of forces = T-mg=Ac
6-4=Ac
2?
 
Closer. The lefthand side is correct, but the righthand side isn't. You can't add up a bunch of forces and then set the result to something that isn't a force. ##a_c## is the centripetal acceleration; it's not a force. It's the ##a## in ##ma## on the righthand side.
 
vela said:
Closer. The lefthand side is correct, but the righthand side isn't. You can't add up a bunch of forces and then set the result to something that isn't a force. ##a_c## is the centripetal acceleration; it's not a force. It's the ##a## in ##ma## on the righthand side.

ah my bad I was correcting it as you answered .
so is it 2=mAc
so its 2 g ? the answer choices only come in a number times g
 
Not quite. You need to figure out what the mass of the object is from its weight.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top