A 22 g ball is fastened to one end of a string
86 cm long and the other end is held fixed at
point O so that the string makes an angle of
35
◦
with the vertical, as in the figure. This
angle remains constant as the ball rotates in
a horizontal circle. The angle
θ
would remain
constant only for a particular speed of the ball
in its circular path.
The acceleration of gravity is 9
Find the velocity
F=Fx+Fy
Two equations:
1.)
Fx=F*Sin35=m*v^2/r
2.)
Fy=F*Cos35-Fg
Fy=F*Cos35-m*g
Solve equation 2 for F:
m*g=F*Cos35
Solving for F:
mg/Cos35=F
Plug 2 into 1 and simplify:
(mg/Cos35)*Sin35=m*v^2/r
mg*Tan35=m*v^2/r
Masses cancel:
g*Tan35=v^2/r
Solve for v:
sqrt[r*(g*Tan35)]=v
Use the last equation, I think.
And plug in the following:
g=9
r=0.86m
Tan35=0.7 *** I think
To find the velocity of the ball, we can use the concept of centripetal acceleration.
Centripetal acceleration is the acceleration directed towards the center of the circular motion. It is given by the equation:
a = v^2 / r
Where:
a is the centripetal acceleration
v is the velocity of the ball
r is the radius of the circular path
In this case, the ball is fastened to one end of the string, and the other end of the string is held fixed. The length of the string is given as 86 cm, which is equivalent to 0.86 meters. This length is the radius of the circular path.
The angle between the string and the vertical is given as 35 degrees. This angle is required for determining the direction of the force as the ball moves in the circular path.
The acceleration due to gravity, represented as g, is given as 9.8 m/s^2.
We know that the acceleration due to gravity acts vertically downward, while the centripetal acceleration acts towards the center of the circular path.
Since the angle between the string and the vertical remains constant, these two accelerations form a right triangle. The vertical component is equal to g, and the horizontal component is equal to the centripetal acceleration.
Using trigonometry, we can determine the horizontal component of the acceleration:
a_horizontal = a * cos(theta)
Substituting the known values:
a_horizontal = g * cos(theta)
a_horizontal = 9.8 * cos(35)
Now, we can substitute this value into the centripetal acceleration equation:
a = a_horizontal = v^2 / r
Thus, we can rearrange the formula to solve for v:
v^2 = a * r
v = sqrt(a * r)
Substituting the known values:
v = sqrt(a * r)
v = sqrt(9.8 * cos(35) * 0.86)
Evaluating this expression will give us the velocity of the ball.