Video Transcript
A ball of mass 1.3 kilograms at the
end of a 1.5-meter string swings in a vertical circle. At its lowest point, the ball is
moving with a speed of 9.5 meters per second. What is the speed of the ball at
the top of its circular path? What is the magnitude of the
tension in the string when the ball is at the top of its circular path? What is the magnitude of the
tension in the string when the ball is at the bottom of its circular path?
In this three-part problem, we can
call the speed of the ball at the top of its path 𝑣 sub 𝑡, the tension at that
point 𝑇 sub 𝑡, and the tension in the string at the bottom of the ball’s path 𝑇
sub 𝑏. Let’s start by sketching out this
vertical circular path. This ball, with a mass of 1.3
kilograms we’ve called 𝑚, moves in a circular vertical arc on a string of length
𝑟, which is 1.5 meters. We know that when the ball is at
the bottom of the circular arc, its speed is given as 9.5 meters per second. Given all this information, we
wanna solve for the speed of the ball when it’s at the top of its arc.
Considering this system as a whole,
since no energy is added to it or taken from it, we can use the conservation of
energy to solve for the velocity of the ball at the top of its arc. We can say that the initial kinetic
plus potential energy of our system is equal to the final kinetic plus potential
energy, where our initial point is at the bottom of the circular arc and the final
point is at the top. Recalling that kinetic energy is
given as one-half an object’s mass times its speed squared and that gravitational
potential energy is 𝑚 times 𝑔 times ℎ, where the acceleration due to gravity, 𝑔,
we’ll treat as exactly 9.8 meters per second squared.
We can write that one-half the
ball’s mass times 𝑣 sub 𝑏 squared is equal to one-half its mass times 𝑣 sub 𝑡
squared plus 𝑚 times 𝑔 times two 𝑟, the radius of the circular arc. We see that the mass 𝑚 cancels
from this expression. And when we rearrange to solve for
𝑣 sub 𝑡, we find it’s equal to the square root of 𝑣 sub 𝑏 squared minus four
times 𝑔 times 𝑟. Since we’re given 𝑣 sub 𝑏 and 𝑟
and know 𝑔 is a constant, we’re ready to plug in and solve for 𝑣 sub 𝑡. Entering this expression on our
calculator, we find that 𝑣 sub 𝑡 is 5.6 meters per second. That’s the speed of the ball at the
top of its circular arc.
Next, we wanna solve for the
tension of the string when the ball is at point 𝑓, at the top of its arc. If we draw a free body diagram
showing the forces acting on the ball at this point, the two forces acting on the
ball at this point are the gravitational force, 𝐹 sub 𝑔, and the tension force, 𝑇
sub 𝑡. Recalling Newton’s second law of
motion that the net force on an object is equal to its mass times its acceleration
and letting motion in the downward direction be positive in our example. We can write that 𝑚𝑔, the
gravitational force, plus 𝑇 sub 𝑡, the tension force from the balls at the top of
its arc, is equal to the mass of the ball times its acceleration, 𝑎.
Because the ball is moving in a
circle, we know that it accelerates centripetally, that is, towards the center of
the circular path, and that the magnitude of this acceleration is equal to 𝑣
squared over 𝑟. So we can write that 𝑚𝑔 plus 𝑇
sub 𝑡 is equal to 𝑚 times 𝑣 sub 𝑡 squared over 𝑟. This expression on the right-hand
side is the centripetal force acting on this ball. We want to rearrange and solve for
𝑇 sub 𝑡. When we do, we see it’s equal to 𝑚
times the quantity 𝑣 sub 𝑡 squared over 𝑟 all minus 𝑔. When we plug in for 𝑚, 𝑣 sub 𝑡,
𝑟, and 𝑔 and enter this expression on our calculator, we find that, to two
significant figures, 𝑇 sub 𝑡 is 15 newtons. That’s the tension in the string
when the ball is at the top of its path.
Lastly, we wanna solve for the
tension in the string when the ball is at the bottom of its arc. To do this, we can once again draw
a free body diagram of the forces on the ball at this point. There are again two forces on the
ball: the tension force, 𝑇 sub 𝑏, and the weight force. But now they point in opposite
directions. We can write that the gravitational
force minus 𝑇 sub 𝑏 is equal to negative 𝑚𝑣𝑏 squared over 𝑟, where the
centripetal acceleration is negative because it points up. When we rearrange this expression
to solve for 𝑇 sub 𝑏, we find it’s equal to 𝑚 times the quantity 𝑣 sub 𝑏
squared over 𝑟 plus 𝑔. When we plug in for these values
and enter this expression on our calculator, we find that 𝑇 sub 𝑏 is 91
newtons. That’s the tension in the string
when the ball is at the bottom of its arc.
