Hi all, I know this may see basic but how would I go about calcualting Impact velovity?
The only information I have is:
Height dropped, height bounced and from here I can calculate the CofR...
But how would I go about finding impact velocity
Calculate the pull on the high bar of a gymnast performing a giant swing. His angular velocity is 2 rads, his mass 80 kg and effector radius of rotation (distance from bar to CofG) is 1 meter
Hi... I am a bit confused... "pull on the bar" would this be the work done on the bar to the person...
An athlete (mass of 50 kg) is moving with a constant velocity of 5ms. Determine the resultant force acting on him.
If it's a constant velocity he is not acclerating? Therefore a = 0 m/s/s; F=ma F=50 X 0 = 0N?
But there has to be a force dosent there? Would the force be equal to the mass...
(Q) What torque is applied to a steering wheel of an automobile when the driver applies a force of 50 Newtons tangent to the circumference of the wheel? Diameter of the steering wheel is 40 cm.
F = 50 N
D (torque arm) = 0.40m
T= Fd = 20 N... but its wrong... any ideas...
So then this indicates that the max velcoity is at the higest point? I though max velcoity is at the lowest point??... but her velocity @ 1m using v= sqrt 2gh also = 4.43 m/s... so is the velcoity then constant?
okay well.. the answer is 4.43 m/s... so v=sqrt (2 x 9.81 x 2) = 4.43m/s... aff yeas on reading the question it say's her velocity at highest point...
Thanks for all your help... I undersatnd how to getthe answer ..but confused how it can be solved without a mass...
rayquesto ... but this would make here velocity at the highest point faster than the lowest point... My undersatdning is velocity is at it/s max when d=1m...
Okay, if PE = 9.81 J @ 2m above the ground... and energy is conserved, @1m above the ground all energy will become KE?; KE=9.81J?
if I plug that into KE=1/2mv(sqr) I still can't comput an answer... I don't have mass
(sorry... this is fustraing the ! out of me lol)... I understand...
Hi all thanks for all you help...
PeterO... the equations 1/2mv(sqr)= 0 = KE @ 2m? and mgh=0 = PE @ 1m?
so.. PE + KE = C
would be: mgh + 1/2mv(sqr) = c?
im really confused... I don't think the question is intended to be this complicated... I think its trying to gte me to think about PE...
All I can think of is the realtionship between energy conservation... So at 2m (max height ) Her PE = max and at 1m (min) her KE=max where her velocity is max and her PE = min/0j)
If I use PE=mgh at 1 m; 0=m x 9.81 x 1 = 9.81 kg...
Then KE at 2m = 1/2mv(sqr) = 0=4.91v(sqr) = 2.2m/s...
The only issue i have using an energy formula is I don't have the mass?
othwise I would use:
PE + KE = C
wt x h x 1/2 mv(sqr) = c
then I would sub in C into:
wt x h x 1/2 mv(sqr) = c, to find V...