Video Transcript
A ball at the end of a rope of
negligible mass moves uniformly along a circular path with a radius of 0.48
meters. The centripetal acceleration of the
ball is 63 meters per second squared. At a point where the rope makes an
angle of 33 degrees above the horizontal, the rope breaks as it moves downwards. At this point, the ball is 1.5
meters vertically above the ground. Find the horizontal distance
between the ball’s position when the rope breaks and its position when it makes
contact with the ground.
This question is asking us about
the horizontal distance a ball travels before it touches the ground after the rope
breaks at this point. We are told the ball’s centripetal
acceleration, as well as the radius of the circular path it travels in. Recall that the equation for
centripetal acceleration, 𝑎 c, is equal to 𝑣 squared over 𝑟, where 𝑣 is the
velocity and 𝑟 is the radius.
Now, when the rope breaks, we can
use this formula to determine the ball’s velocity at that moment, though after this
happens, the ball will follow a parabolic arc downwards. So it will have both horizontal and
vertical velocity. This means we also have to use the
equations for parabolic motion.
Now, let’s note that centripetal
acceleration does not change the magnitude of the velocity, only the direction. Thus, when the string breaks, the
ball comes out with a certain speed 𝑣, which we can determine by using the equation
for centripetal force. Solving for 𝑣 requires us to first
multiply both sides by 𝑟, which cancels 𝑟 on the right side. Then, we take the square root of
both sides to cancel the squared term for 𝑣.
Okay, now, we are given the
centripetal acceleration as 63 meters per second squared and the radius as 0.48
meters. Substituting these values into the
equation and calculating gives us a value of velocity of 5.5 meters per second. Since this velocity comes from a
centripetal acceleration, the velocity must be tangential to the circular path it
was traveling in and is therefore tangential at point 𝐴, where the rope breaks.
When the rope breaks, the force of
the tension is zero. So the only force on the ball at
that time will just be gravity. That means from here, we can think
of this ball as just having a typical parabolic trajectory, with an initial velocity
of 5.5 meters per second, height of 1.5 meters, and acceleration due to gravity of
9.8 meters per second squared. We can draw the approximate
trajectory that the ball will travel until it reaches the ground.
Now, we need to know the magnitudes
of the horizontal and vertical velocities. We can calculate them using the
angle we are given from the problem of 33 degrees. However, we won’t be using the
obvious triangle we are given, but rather this triangle here, where the hypotenuse
is the direction of velocity. The angle of 33 degrees is also the
same here though, since these are the same triangles. Now then, using this triangle, the
velocity times the cos of 33 degrees will give the vertical component of velocity,
𝑣 𝑦, while the sin of 33 degrees will give the horizontal component, 𝑣 𝑥.
Substituting in our value of
velocity, 5.5 meters per second, and calculating through, we get that the horizontal
velocity is 2.995 meters per second and the vertical velocity is 4.612 meters per
second. We then set our vertical velocity
to be negative, since it is in the downwards direction here.
Finally, let’s recall the equations
of parabolic motion. The horizontal position, 𝑥, is
equal to the initial horizontal position, 𝑥 naught, plus the initial horizontal
velocity, which is 𝑣 𝑥 in this case, times the time the ball is in the air,
𝑡. The vertical position 𝑦 is given
by the initial vertical position 𝑦 naught plus the initial vertical velocity 𝑣 𝑦
times the time plus one-half the acceleration due to gravity 𝑔 times 𝑡
squared. In this case, the gravity will be
negative, since it is applying a downward force.
Now, we want to find the total
horizontal distance the ball travels from when the rope breaks, which will just be
the horizontal position. Since the initial horizontal
distance, 𝑥 naught, is zero, and the horizontal velocity won’t change, the equation
will just look like 𝑥 equals 𝑣 𝑥 𝑡. We already know 𝑣 𝑥, so all we
need now is 𝑡, which we can obtain from the equation for vertical position. We want to find when the ball hits
the ground. So we should set our vertical
position to zero, leaving us with zero equals 𝑦 naught plus 𝑣 𝑦 𝑡 plus one-half
𝑔𝑡 squared. Our initial vertical position is
the height, which is 1.5 meters. The initial vertical velocity is
negative 4.62 meters per second. And the gravitational acceleration
is negative 9.8 meters per second squared. So, let’s substitute these data
in.
Now, this should start to look
familiar. We’re going to need to use the
quadratic formula to find 𝑡, where the variable, in this case 𝑡, is equal to
negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all divided
by two 𝑎, where 𝑎 is the 𝑡 squared term, negative one-half 9.8, which is just
negative 4.9. 𝑏 is the 𝑡-term, negative
4.612. And 𝑐 is the constant term,
1.5. Substituting in these values and
calculating, we get negative 0.4706 seconds plus or minus negative 0.7263
seconds.
Time must be greater than zero
because it cannot be negative. So, the time must be negative
0.4706 seconds minus negative 0.7263 seconds, which is equal to 0.2557 seconds. We can now substitute this into our
equation for total horizontal distance traveled, along with the horizontal
velocity. 2.995 meters per second times
0.2557 seconds is equal to 0.767 meters, which, rounded to two decimal places, is
0.77 meters.
Therefore, the horizontal distance
the ball travels from when the rope breaks to when it touches the ground is 0.77
meters.
