2 M3 Circular Motion Qs (Chap 4E) Watch

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Report Thread starter 13 years ago
QS 3

A pendulum bob has mass 0.2kg. It is attached to one end of a light rod of length 2m. The rod is free to rotate in a vertical plane about an axis through the other end O. Given that the pemdulum swings through 60 degrees on either side of the vertical, calcualte the speed of the bob at the lowest point of its path and the tension in the rod when the rod makes an angle of 30 degrees with the downward vertical.

First part doesnt need help but in 2nd part i dont know how the components resolve and how the height differs of the ball when it travels in its path when the angle changes. ( a diagram wuld be appreciated)

QS 6

A bead B of mass m is threaded on a smooth circular wire of radius a and centre O which is fixed in a vertical plane. B is released from rest at the point where OB maeks an angle of 30 degrees with the upward vertical. Given that the speed of B at the lowest point of its path is v, find v^2 in terms of a and g. b) find in terms of m and g the magnitude and direction of the reaction of the wire on the bead when OB makes an angle of 60 degrees with the upward vertical.

thank you very much.
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Report 13 years ago
The vertical distance of the bob from O when it makes an angle 30 with the vertical equals 2cos30. Similarly, when it makes an angle 60, the distance is 2cos60. So the difference in 'height' betewen the bob at 60 and the bob at 30 is:
h = 2co30 - 2cos60

Conservation of energy between those two positions:
0.5mv² = mgh
v² = 2gh = 4*9.8*(cos30-cos60) =~ 14.3 m/s

When the bob makes an angle 30 with the vertical, we resolve to get:
T - mgcos30 = mv²/r
T = mgcos30 + mv²/2
= 0.2*9.8*0.5*rt3 + (0.2*14.3)/2
= 3.13N

Initially, the height of B above O is acos30. So, if we take the lowest point to be our reference level:
PE at B = mgh = mg(a+acos30)

At the lowest level:
KE = 0.5mv²

Conservation of energy:
0.5mv² = mg(a+acos30)
v² = 2ga(1+cos30) = ga(2+rt3)

When OB makes an angle 30, the vertical height of the bead above O is acos30.
When OB makes an angle 30, the vertical height of the bead above O is acos60.
So, the difference in height is:
h = a(cos30-cos60)

Conservation of energy, as before, gives:
v² = 0.732ga

And resolving gives:
mgcos60 + R = mv²/r
R = mv²/r - mgcos60
= m(0.732g - 0.5g)
= 0.232mg N towards O.
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