Circular Motion?

Question

A 920 g rock is whirled in a horizontal circle at the end of a 1.5 m-long string.

If the breaking strength of the string is 120 N, what is the maximum allowable speed of the rock?

At this maximum speed, what angle does the string make with the horizontal?

My Work:

I have successfully found the max speed to be 14 m/s (which is the correct answer) and I assumed that the angle it makes with the horizontal at the speed 14 m/s was zero since I used zero for theta to get the speed in the first place but it was incorrect.

I tried using 180 and 360 and a whole bunch of crazy unrelated thetha just for the heck of it and none of em seems to work.

Am I missing something big here?

Thanks

Answer 1

The rock has only 2 forces acting on it. Gravity and tension. Since you want the rock to have uniform circular motion, the net force must be exactly horizontal towards the center of rotation. The force of gravity is down so the tension force must also have a component upwards to counteract gravity.

The component of the tension force in the vertical direction is 120N * sin(theta) (where theta is the angle between the string and horizontal)
Set this equal to the gravitational force and solve for theta.

You said your answer of 14m/s with using theta = 0 came out right. The actual speed should be a little less than what you got when you take gravity into account because gravity adds a little to the tension. So less speed is needed to get the same tension. The speed will probably change very little though. Once you round it to two significant digits it will most likely still be 14 m/s.

Once you figure out theta, try finding the speed again. This time the force towards the center will be the sum of the tension force and gravitational force. The net force towards the center will be slightly less.

Edit:
120N sin(theta) = 0.92 kg 9.8m/s^2
sin(theta) = 0.075133333
theta = 0.075204 radians or 4.30482 degrees.

r = 1.5m cos(theta) = 1.49576m
f(net) = 120N cos(theta) = 119.66082N horizontally
|f(net)| = 0.92 kg * v^2/ 1.49576m = 119.66082N
v^2 = 194.5477 m^2/s^2
v = 13.9480 m/s

Answer 2

The only way I can interpret the question is to assume that the vector sum of the centrifugal force (CF) and gravity force (W) results in an angle ø downward from the horizontal

CF = m*v^2/r

W= m*g

ø = arctan(g*r/v^2)

To properly take the angle into account, you would have to compute the string tension as

T^2= (m*g)^2 + (m*v^2/r)^2

(m*v^2/r)^2 = T^2 - (m*g)^2

m*v^2/r = √[T^2 - (m*g)^2]

v^2 = (r/m)*√[T^2 - (m*g)^2]

v = √[(r/m)*√[T^2 - (m*g)^2]]

This comes out v = 13.968 m/sec

and ø = 0.0752 radian or 4.31º

Answer 3

The angle of the string with the horizontal when it breaks must be
arctan((mg)/(mv^2/r) = arctan(rg/v^2)
F = √((mg)^2 + (mv^2/r)^2) = m√(g^2 + v^4/r^2)
g^2 + v^4/r^2 = (F/m)^2
v^4/r^2 = (F/m)^2 - g^2
v^4 = r^2((F/m)^2 - g^2)
v^2 = r√((F/m)^2 - g^2)
v = √(1.5√((120/0.920)^2 - (9.80665)^2
v ≈ 13.96776 m/s ≈ 14.0 m/s

θ = arctan(rg/v^2)
θ ≈ arctan(1.5*9.80665/195.0984)
θ ≈ arctan(0.07539772)
θ ≈ 4.311812°

Answer 4

I don't know why the angle would make a difference but try 45 degrees