# Mechanics help

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#1
A past paper question in a multiple choice physics paper I'm working on says:

"A simple pendulum consists of a mass of 0.200kg on a light inextensible string of length 1.20m. The mass is displaced sideways until it has risen 0.050m from its equilibrium position and then it is released. What is the tension in the string as the mass passes through its lowest position?"

A: 0.363N
B: 1.80N
C: 1.96N
D: 2.13N

I got B by equating the tension T in the string with mg - mv^2/r, and calculating v^2 using GPE lost = KE gained.

However the correct answer is D. Could someone help me understand why this is the right answer and where I've gone wrong? Thank you!
0
7 months ago
#2
Why are you subtracting?

(Original post by asliceofpi)
A past paper question in a multiple choice physics paper I'm working on says:

"A simple pendulum consists of a mass of 0.200kg on a light inextensible string of length 1.20m. The mass is displaced sideways until it has risen 0.050m from its equilibrium position and then it is released. What is the tension in the string as the mass passes through its lowest position?"

A: 0.363N
B: 1.80N
C: 1.96N
D: 2.13N

I got B by equating the tension T in the string with mg - mv^2/r, and calculating v^2 using GPE lost = KE gained.

However the correct answer is D. Could someone help me understand why this is the right answer and where I've gone wrong? Thank you!
tension in the string when the mass is stationary at equilibrium position is 1.96N ... when the mass passes through the equilibrium at speed there must be an additional centripetal force causing it to accelerate towards the centre (i.e. upwards)

you could go straight for D without any more calculation because it's the only option with a tension greater than 1.96N
0
7 months ago
#3
(Original post by asliceofpi)
A past paper question in a multiple choice physics paper I'm working on says:

"A simple pendulum consists of a mass of 0.200kg on a light inextensible string of length 1.20m. The mass is displaced sideways until it has risen 0.050m from its equilibrium position and then it is released. What is the tension in the string as the mass passes through its lowest position?"

A: 0.363N
B: 1.80N
C: 1.96N
D: 2.13N

I got B by equating the tension T in the string with mg - mv^2/r, and calculating v^2 using GPE lost = KE gained.

However the correct answer is D. Could someone help me understand why this is the right answer and where I've gone wrong? Thank you!

I hope this is still relevant. I can't comment on your results once you haven't post your workings. However you need to observe that tension in this question was the sum of Weight plus the centripetal force ( mV^2/r) . You confused yourself, there is in fact a relationship that follows the conservation of energy and therefore when the pendulum is at its maximum displacement the system has maximum potential energy, however zero kinetic energy. The opposite is observed when the pendulum is at its equilibrium position. Therefore you can find the velocity^2 by setting mgh=1/2mv^2. the m's cancel and all you need to do is to plug the numbers h=0.05m and g=9.81. Solve for V^2 and substitute onto the centripetal force equation, r is the radius which you take as the length of the string. Set T=W+centripetal force and there you have it. I have attache the working if you want it.

Attachment 882860
Last edited by ThiagoBrigido; 7 months ago
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