Physics Question Help?

Question

A 21g ball is fastened to one end of a string 68cm long and the other end is fixed to a point on a ceiling that makes an angle of 28 degrees with the vertical. This angle remains constant as the ball rotates in a horizontal circle. The angle theta would remain constant only for a particular speed of the ball in a circular path. The accleration og gravity is 9.8m/s2. FInd this speed. Answer in m/s

Answer 1

i dont no the speed

Answer 2

Since the ball moves in a circle, the net force on the ball is directed radially inward in the plane of the circle, with magnitude v^2/r, where v is the speed of the ball and r is the radius of the circle. You can calculate r from the length and angle of the string, right? (A force that causes circular motion is called a centripetal force.)

This gives the net force as m*v^2/r, also directed radially inward. This net force must be the vector sum of all external forces, of which there are two: the downward force of gravity, and the force from the tension (T) in the string.

The net vertical force is zero, so the vertical component of the string tension force must be mg. (Why?)

This lets you compute the tension T of the string in terms of m and v (which are given) and r (which you calculated above).

Next, the horizontal component of the string tension force is the only horizontal force, so it must be equal to the centripetal force, m*v^2/r above. Express this as an equation and solve for v in terms of m (given), and the calculated values of T and r.

Hint the components of string tension are T*sin(23) and T*cos(23). I'm sure you can figure out which is horizontal.

Hint 2: Mass should cancel out in the final answer. See if you can see why this should be so as you solve the problem.

Happy Solving!

Answer 3

The poster above, husoski is correct. I would just like to elaborate on the details.

Centripetal Force:
F = (mv^2)/r

r = (68 cm)(sine 28 degrees)
r = (68)(0.4695) = 31.926 cm ---> radius

g = 9.8 m/s2 = 980 cm/s2 ---> gravitational acceleration

W = mg = (21 g)(980 cm/s2)
W = 20580 g-cm/s2 ---> weight

Tv = T [sine (90 - 28) degrees] ----> Vertical. component of string tension (equals the weight W)

note: if the string makes an angle of 28 degrees with the vertical axis, then it also makes an angle of (90 - 28) degrees with the horizontal axis (rule on complementary angles of a right triangle). The rules on trigonometric functions will also be complementary (e.g. sine 28 deg = cosine (90 - 28) deg, cosine 28 deg = sine (90 - 28) deg, etc.).

Tv = W = T (cosine 28 degrees)
T = W/(cosine 28 degrees) ---> string tension

F = Th = T [cos (90 - 28) degrees] ----> Horiz. component of string tension
F = Th = T (sine 28 degrees)
F = Th = [W/(cosine 28 degrees)] (sine 28 degrees)
F = Th = W (sine 28 degrees)/(cosine 28 degrees))

F = Th = W (tan 28 degrees) = (20580)(0.532)
F = 10948.56 ---> centripetal force

F = (mv^2)/r
v^2 = (Fr)/m
v^2 = (10948.56)(31.926)/(21)
v^2 = 16644.939
v = 129.015 cm/sec ---> answer (speed)

Answer 4

calculate the circumfrence of the circle using trig

tan28 = r/68
68xtan28 =r
2pi r =C
C= distance
speed = d/t
then it falls down a bit im still working on it