A 0.20-kg ball on a string is whirled on a vertical circle at a constant speed. When the ball is at the three o'clock position, the string tension is 16 N. Find the tensions in the string when the ball is at twelve o'clock and at the six o'clock positions.
* Physics - bobpursley, Tuesday, October 6, 2009 at 4:52pm
at three oclock, Tension= mv^2/r
then at 12 oclock,
tension= mv^2/r-mg
and at six...
tension= mv^2/r + mg
How do I solve for the r and the v?
YOu don't need it.
you know mv^2/r=16N
Just sub some numbers in to make it work. For example... tension/mass (16/.2)=80. So 80 is equal to rv^2. so make v=10 m/s and r =.8 m so everything will work out. It stays the same throughout so you can 'make up' the numbers
To solve for the radius (r) and the velocity (v) in this problem, you'll need to use the given information about the tension in the string at the three o'clock position.
First, let's start with the equation for tension at the three o'clock position: Tension = mv^2/r
Given:
Mass of the ball (m) = 0.20 kg
Tension at the three o'clock position = 16 N
Using this information, we can rearrange the equation to solve for v^2/r:
16 N = (0.20 kg) * v^2 / r
Now, to determine the tension at the twelve o'clock position, we'll use the equation: Tension = mv^2/r - mg
To solve for the tension at twelve o'clock, we need to know the value of g, which is the acceleration due to gravity. Assuming that this is on Earth, g is approximately 9.8 m/s^2.
Similarly, to find the tension at the six o'clock position, we'll use the equation: Tension = mv^2/r + mg.
By substituting the values of mass, tension, and gravity into these equations, you can calculate the tensions at the twelve o'clock and six o'clock positions.
Keep in mind that in this problem, you were asked to find the tensions when the ball is at twelve o'clock and six o'clock, so you don't need to solve for the radius or the velocity.