A boy is whirling around a stone tied at the end of a 0.71 m long string at constant angular speed of 4 revolutions per second in a plane perpendicular to the ground. The string breaks when it is making an angle of 54 degrees above the horizontal and the stone is on the way up. The stone is 0.93 m above the ground when this happens. How long does the stone take to hit the ground after the string breaks?
"a stone tied at the end of a 0.71 m long string"
r=0.71 m
"4 revolutions per second"
ω=4*2π radians/s
Tangential speed, v0
=rω
"The stone is 0.93 m above the ground"
Distance above ground, h
= 0.93
"an angle of 54 degrees above the horizontal"
θ=54°
Vertical component of initial velocity, vv
=v0sinθ
Let t=time it takes to hit ground,
S=vv*t+(1/2)gt²
-0.93 = vv*t+(1/2)(-9.8)t²
Solve for t.
I get -0.051 sec. or 4 sec. approx.
We reject -0.051 because t>0.
I tried doing the problem your way, but I got confused at the end.
For the equation S = w*t + (1/2)(-9.8)t**2, what was the value of w?
When I tried using w = v0 (17.844m/s), I got the same answers as yours
But when I tried using w = v0sin54 (14.4363m/s), I got different answers. I'm not sure which way is right. I'm leaning toward the latter one.
You're quite right.
I have calculated for 5 revolutions per second. That is why they don't match your answers, which are on the right track.
I get now -0.06 and 3 seconds, and -0.06 is to be rejected.
What do you get for your answers?
Sorry for the inconvenience.
Yes, I also got -0.06 and 3.00 seconds, so obviously the answer is 3.00 seconds.
Thanks for your help :)
You're welcome!
To find the time it takes for the stone to hit the ground after the string breaks, we can use a few key concepts of circular motion and projectile motion.
First, let's find the initial velocity of the stone when the string breaks. We know that the stone is 0.93 m above the ground and making an angle of 54 degrees with respect to the horizontal. This forms a right-angle triangle with the vertical distance and the horizontal distance.
Using the trigonometric relation, we can calculate the vertical distance traveled by the stone:
Vertical distance = (0.93 m) * sin(54 degrees)
Vertical distance ≈ 0.757 m
Since the stone is moving at constant angular speed, we can relate the angular speed to the linear speed using the formula:
Linear speed = Angular speed * Radius
Given that the radius is 0.71 m and the angular speed is 4 revolutions per second, we can calculate the linear speed:
Linear speed = (4 revolutions/sec) * (2π rad/revolution) * (0.71 m)
Linear speed ≈ 8.92 m/s
Now, using the vertical velocity component, we can determine the time it takes for the stone to reach the ground. Since we know the initial vertical velocity is 0 (the stone is momentarily at rest when the string breaks), we can use the equation for vertical motion:
Vertical distance = Initial vertical velocity * time + (1/2) * acceleration * time^2
Acceleration due to gravity is acting in the downward direction, so we'll use a value of -9.8 m/s². Rearranging the equation and plugging in the values, we get:
0.757 m = (1/2) * (-9.8 m/s²) * time^2
Simplifying the equation, we have:
0.757 m = -4.9 m/s² * time^2
Rearranging the equation again, we find:
time^2 = (0.757 m) / (4.9 m/s²)
time^2 ≈ 0.1546 s
Taking the square root of both sides, we get:
time ≈ 0.3938 s
Therefore, the stone takes approximately 0.3938 seconds to hit the ground after the string breaks.