Ask your own question, for FREE!
Physics 11 Online
OpenStudy (anonymous):

A pendulum is used in a large clock. The pendulum has a mass of 2 kg. If the pendulum is moving at a speed of 4.1 m/s when it reaches the lowest point on its path, what is the maximum height of the pendulum? A. 0.86 m B. 0.63 m C. 0.51 m D. 0.75 m

OpenStudy (roadjester):

how far along are you in physics? have you learned about oscillatory motion? or have you only learned about momentum?

OpenStudy (anonymous):

i don't really know that much..

OpenStudy (roadjester):

do you know what simple harmonic motion is?

OpenStudy (roadjester):

@LastDayWork I'm actually not sure how to approach this problem. Pendulums are typically part of oscillatory motion, but as of right now @marius.roine has only asked me about collisions which is part of linear momentum. Because of that I've considered using an angular momentum approach but i'm stuck

OpenStudy (lastdaywork):

Well, I can't see any correlation with collision here. Simply conserving the mechanical energy will give the answer.

OpenStudy (roadjester):

\[E_i = E_f\] but are you referring to kinetic or potential energy?

OpenStudy (roadjester):

Only mass and velocity are known, that's not much to go by Then again, I'm a bit rusty in the energy approach in mechanics

OpenStudy (lastdaywork):

(1/2)mv^2 = mgh We can find 'h' from the above equation.. We could also use angular approach; but I prefer doing things in a lazy way ;)

OpenStudy (roadjester):

then what's the point of knowing mass if it just cancels out? or is that for the angular approach I suggested?

OpenStudy (lastdaywork):

Mass will cancel out in the angular approach too...try it :)

OpenStudy (roadjester):

I actually don't even know how to do it that way. that's why i asked you for help :( it's kinda sad really, I know it's possible, I don't know the how

OpenStudy (lastdaywork):

Only mg can give torque as Tension would always be directed towards the axis of rotation. The approach would be a little messy as the component of mg, which gives torque, will be changing with time.

OpenStudy (anonymous):

(1/2) m v^2 = m g h sure is a clean solution method

OpenStudy (anonymous):

what will the answer be then @LastDayWork @roadjester

OpenStudy (lastdaywork):

Use - (1/2)mv^2 = mgh and find 'h'

OpenStudy (anonymous):

so 1/2 x 2 x 4.1^2 = 16.81 @LastDayWork

OpenStudy (lastdaywork):

Yea..now equate the expression to RHS..

OpenStudy (anonymous):

how can I do that

OpenStudy (lastdaywork):

Do you know how to solve equations ??

OpenStudy (anonymous):

yes, but not this one

OpenStudy (lastdaywork):

Use - h = (1/2) v^2 /g

OpenStudy (lastdaywork):

where g is the acceleration due to gravity.

OpenStudy (anonymous):

so it wil be A

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!