T(top ) = m(v²/R – g ) =796 N
T(bottom ) = m(v²/R + g) = 894 N
A ball having a mass of 5 kg is attached to a string 1 m long and is whirled in a vertical circle at a constant speed of 13 m/s.
(a) Determine the tension in the string when the ball is at the top of the circle.
(b) Determine the tension in the string when the ball is at the bottom of the circle.
The answers are supposed to be
a) 796 N
b) 894 N
How do I get these answers? Thank you in advance.
6 answers
When I use the first equation I get -90.8 N. What am I doing wrong?
I'm putting this is my calculator:
5(169/(.5-9.81)) = -90.76262084 N
I'm putting this is my calculator:
5(169/(.5-9.81)) = -90.76262084 N
T(top)= 5•(13²/1-9.8)=5•(169-9.8)=
=5•159.2=796 N.
T(bottom) = 5•(13²/1+9.8)=5•(169+9.8)=
=5•178.8=894 N
=5•159.2=796 N.
T(bottom) = 5•(13²/1+9.8)=5•(169+9.8)=
=5•178.8=894 N
I see what I was doing wrong now. I was putting the parenthesis in the wrong places. Thank you!
Why does the force done by gravity change from positive to negative?
Really we have (according to Newton’s 2 law)
(positive direction of y-axis - downwards)
for the top point:
ma =T+mg => mv²/2=T+mg,
T= mv²/2 - mg
for the bottom point:
- ma = - T+mg => - mv²/2= -T+mg ,
T= mv²/2 + mg
(positive direction of y-axis - downwards)
for the top point:
ma =T+mg => mv²/2=T+mg,
T= mv²/2 - mg
for the bottom point:
- ma = - T+mg => - mv²/2= -T+mg ,
T= mv²/2 + mg