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Question about centripetal acceleration.

Dhamnekar Winod

Active member
Nov 17, 2018
149
Hi,

A mother pushes her child on a swing so that his speed is 9.00 m/s at the lowest point of his path. The swing is suspended 2.00 m above the child’s center of mass.

(a) What is the magnitude of the centripetal acceleration of the child at the low point?

(b) What is the magnitude of the force the child exerts on the seat if his mass is 18.0 kg?

(c) What is unreasonable about these results?

(d) Which premises are unreasonable or inconsistent?

Answer:- (a) $\frac{(9.00 m/s)^2}{2.00 m}=40.5 m/s^2$

(b) Answer given is 905 N. But my answer is $40.5 m/s^2 \times 18.00kg =729 N$ What is wrong with my answer?

How would you answer question (c) and (d)?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
Answer:- (a) $\frac{(9.00 m/s)^2}{2.00 m}=40.5 m/s^2$

(b) Answer given is 905 N. But my answer is $40.5 m/s^2 \times 18.00kg =729 N$ What is wrong with my answer?
There are two forces acting on the child: the normal force from the swing, and the gravitational force.
They are opposite and they must add up to the centripetal force.
In other words, you still need to add the gravitational force.

How would you answer question (c) and (d)?
The acceleration $40.5\,m/s^2$ is about 4 times the acceleration due to gravity. Add gravity itself and we have 5 times. That seems like a lot to subject a child to.
The maximum speed that the child can realistically have, is if we bring the swing up all the way to the side and 2.00 meters up, and then let it swing down.
What will the speed of the child then be at the lowest point?
 

Dhamnekar Winod

Active member
Nov 17, 2018
149
Hi,
How to compute the maximum speed that the child can realistically have at the lowest point of his path??
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
Hi,
How to compute the maximum speed that the child can realistically have at the lowest point of his path??
Easiest is to apply conservation of energy.
The added gravitational energy is $mgh$ with $h=2.00\,m$, which must be equal to the kinetic energy $\frac 12 m v^2$ at the lowest point (disregarding friction).
That is:
$$mgh = \frac 12 mv^2\implies v=\sqrt{2gh}$$
 
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