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A ball having a mass of 4 kg is attached to a string 1 m long and is whirled in a vertical circle at a constant speed of 23 m/s...Asked by Camilla
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.
(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.
Answers
Answered by
Elena
T(top ) = m(v²/R – g ) =796 N
T(bottom ) = m(v²/R + g) = 894 N
T(bottom ) = m(v²/R + g) = 894 N
Answered by
Camilla
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
Answered by
Elena
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
Answered by
Camilla
I see what I was doing wrong now. I was putting the parenthesis in the wrong places. Thank you!
Answered by
Camilla
Why does the force done by gravity change from positive to negative?
Answered by
Elena
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
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