The Student Room Group

M3: Motion in a vertical circle

(Paper attached)

Two questions:

Firstly, with vertical SHM (particle on string). If T=0, then do you assume it's at natural length? (Jan 07 Q7c - x value )


Also, Question 4 part c) - why isn't the component of velocity cos not sin? :s-smilie:

Thank you :smile:
Reply 1
Original post by lekha2611
(Paper attached)

Two questions:

Firstly, with vertical SHM (particle on string). If T=0, then do you assume it's at natural length? (Jan 07 Q7c - x value )


Also, Question 4 part c) - why isn't the component of velocity cos not sin? :s-smilie:

Thank you :smile:


For the first question;

EDIT: Are you talking about time or tension in the first question? If its tension then yes at natural length there is no tension in the string if it is the vertical plane the particle is then acting under gravity alone.
If talking about time. The particle is released from point B initially so at time t=0 the particle is at B.

Second question;

It's hard to explain without a diagram but I assume you know the velocity acts through the line of the tangent to the circle. If you carefully draw out a diagram and make sure you get your angles right you will see its sin not cos
(edited 11 years ago)
Reply 2
Original post by TheJ0ker
For the first question;

EDIT: Are you talking about time or tension in the first question? If its tension then yes at natural length there is no tension in the string if it is the vertical plane the particle is then acting under gravity alone.
If talking about time. The particle is released from point B initially so at time t=0 the particle is at B.

Second question;

It's hard to explain without a diagram but I assume you know the velocity acts through the line of the tangent to the circle. If you carefully draw out a diagram and make sure you get your angles right you will see its sin not cos


1 - I was talking about tension sorry!
2 - see attachment - is this the correct component?

Thanks :smile:
Reply 3
Original post by lekha2611
1 - I was talking about tension sorry!
2 - see attachment - is this the correct component?

Thanks :smile:


The angle is correct but the hypotenuse of the new triangle you have drawn is not equal to v so it's not vcos theta. I will try and draw a diagram now hold on
Reply 4
Original post by lekha2611
1 - I was talking about tension sorry!
2 - see attachment - is this the correct component?

Thanks :smile:


TSR help 1.png
Reply 5
Original post by TheJ0ker
TSR help 1.png


Ah I see it now - thank you! :biggrin:

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