- #1
Sdarcy
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Okay, I have given this a go but its been 2 years since I've done any dynamics so I think I've done something stupid...
A ball is attached horizontally by a string of length L to a central point C. The mass, m, of the ball is 4.775kg. It is released from rest and allowed to swing downwards. What is the tension in the string (in N) when the ball has fallen through 45 degrees.
This is what I've done so far:
[tex]\sum[/tex]F[tex]_{}n[/tex] = ma[tex]_{}n[/tex]
T- m sin[tex]\alpha[/tex] = m/g (v[tex]^{}2[/tex]/L)
T = m(sin[tex]\alpha[/tex] + v[tex]^{}2[/tex]/gL)
[tex]\sum[/tex]F[tex]_{}t[/tex] = ma[tex]_{}t[/tex]
m cos[tex]\alpha[/tex] = m/g a[tex]_{}t[/tex]
a[tex]_{}t[/tex] = g cos[tex]\alpha[/tex]
vdv = a[tex]_{}[/tex]t ds
ds = L d[tex]\alpha[/tex]
vdv = a[tex]_{}[/tex]t L [tex]\alpha[/tex]
vdv = g L cos[tex]\alpha[/tex] d[tex]\alpha[/tex]
then integrate that equation
v = [tex]\sqrt{}[/tex]2gl(sin[tex]\alpha[/tex] - sin[tex]\alpha[/tex]0)
T = m(3 sin [tex]\alpha[/tex] - 2 sin [tex]\alpha[/tex]0)
= 4.775[3sin(45) - 2sin(0)]
= 10.13N
Which is apparently very, very wrong.
Not sure what I've stuffed up, but help would REALLY be appreciated. Thanks.
A ball is attached horizontally by a string of length L to a central point C. The mass, m, of the ball is 4.775kg. It is released from rest and allowed to swing downwards. What is the tension in the string (in N) when the ball has fallen through 45 degrees.
This is what I've done so far:
[tex]\sum[/tex]F[tex]_{}n[/tex] = ma[tex]_{}n[/tex]
T- m sin[tex]\alpha[/tex] = m/g (v[tex]^{}2[/tex]/L)
T = m(sin[tex]\alpha[/tex] + v[tex]^{}2[/tex]/gL)
[tex]\sum[/tex]F[tex]_{}t[/tex] = ma[tex]_{}t[/tex]
m cos[tex]\alpha[/tex] = m/g a[tex]_{}t[/tex]
a[tex]_{}t[/tex] = g cos[tex]\alpha[/tex]
vdv = a[tex]_{}[/tex]t ds
ds = L d[tex]\alpha[/tex]
vdv = a[tex]_{}[/tex]t L [tex]\alpha[/tex]
vdv = g L cos[tex]\alpha[/tex] d[tex]\alpha[/tex]
then integrate that equation
v = [tex]\sqrt{}[/tex]2gl(sin[tex]\alpha[/tex] - sin[tex]\alpha[/tex]0)
T = m(3 sin [tex]\alpha[/tex] - 2 sin [tex]\alpha[/tex]0)
= 4.775[3sin(45) - 2sin(0)]
= 10.13N
Which is apparently very, very wrong.
Not sure what I've stuffed up, but help would REALLY be appreciated. Thanks.