A light string of length of 56.0 cm connects two small square blocks, each with a mass of 1.50 kg. The system is placed on a
slippery (frictionless) sheet of horizontal ice and spun so the two blocks rotate uniformly about their common
center of mass,
which itself does not move. They are supposed to rotate with a period of 0.750 s. If the string can exert a force of only 100
N
before it breaks, determine whether this string will work.
All i know is the answer is 29.5N but i dont know how they got that answer. help????
r = 0.56/2 = 0.28 meters
period = .75 s
so the block goes 2 pi r meters in 0.75 seconds
v = 2 pi *.28 / 0.75 = 2.35 meters/second
F = mass * centripetal acceleration
= m v^2/r
= 1.5 * 2.35^2/.28 = 29.5 Newtons
Thank you!
You are welcome.
To determine whether the string will work, we need to calculate the tension in the string.
First, let's find the rotational speed or angular velocity (ω) of the system. The period (T) is the time taken for one complete rotation, so we can find ω using the formula:
ω = 2π / T
ω = 2π / 0.750 s
ω ≈ 8.3776 rad/s (rounded to four decimal places)
Next, we need to find the centripetal force acting on each block. The centripetal force (Fc) is given by the formula:
Fc = mω²R
where m is the mass of each block and R is the radius of rotation (half the length of the string).
Given:
m = 1.50 kg (for each block)
R = length of string / 2 = 56.0 cm / 2 = 28.0 cm = 0.28 m (converted to meters)
Substituting the values, we get:
Fc = (1.50 kg)(8.3776 rad/s)²(0.28 m)
Fc ≈ 11.614 N (rounded to three decimal places)
Since there are two blocks, the total centripetal force on the string is twice this value:
Total Fc = 2(11.614 N)
Total Fc ≈ 23.228 N (rounded to three decimal places)
Now, we need to check if the tension in the string is less than the maximum force it can exert before breaking. If the tension (T) in the string is less than 100 N, then the string will work.
Tension in the string (T) = Total Fc
T ≈ 23.228 N
Comparing the tension (23.228 N) with the maximum force the string can exert before breaking (100 N), we see that T < 100 N. Therefore, the string will work.
Note: The answer you mentioned (29.5 N) seems to be incorrect based on the calculations I provided.