Estimate the force a person must exert on a string attached to a 0.200 kg ball to make the ball revolve in a circle when the length of the string is 0.600 m. The ball makes 1.40 revolutions per second. Do not ignore the weight of the ball. In particular, find the magnitude of FT, and the angle phi it makes with the horizontal. [Hint: Set the horizontal component of FT equal to maR; also, since there is no vertical motion, what can you say about the vertical component of FT?]
FT = ? N
Angle = ? degrees
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MY WORK:
FTy = Fg = (9.8)(.2) = 1.96 N
v = 2 rps * 2π(0.6) = 1.68π m/s
FTx = Fc = (mv^2)/r
FTx = [(.2)(1.68π)^2]/(0.6)
FTx = 9.28532 N
FT^2 = FTx^2 + FTy^2
FT = root(90.0587675)
FT = 9.4899 [FINAL ANSWER]
FTy = FTsin(angle)
sin^-1(1.96)/(9.4899) = angle
Angle = 11.9 degrees [FINAL ANSWER]
...so, what exactly am I doing wrong? The Webassign grader told me both were incorrect, but I've worked this over a million times and it's always come out the same.
Where is the 90.06 coming from?
Let's go step-by-step to solve this problem again and see where the error might be:
1. Determine the vertical component of the tension force (FTy):
FTy = Fg (gravity) = mg = (0.2 kg)(9.8 m/s^2) = 1.96 N
2. Determine the tangential speed of the ball (v):
The ball makes 1.4 revolutions per second, which is equivalent to 2π(1.4) radians per second.
v = rω = (0.6 m)(2π)(1.4 rad/s) = 5.28 m/s
3. Determine the horizontal component of the tension force (FTx):
FTx provides the centripetal force required to keep the ball moving in a circle.
FTx = mv²/r = (0.2 kg)(5.28 m/s)² / 0.6 m = 18.432 N
4. Calculate the magnitude of the total tension force (FT):
FT² = FTx² + FTy²
FT = sqrt(FTx² + FTy²) = sqrt((18.432 N)² + (1.96 N)²) = sqrt(340.41216 N²) = 18.45 N
Now let's calculate the angle phi the force FT makes with the horizontal:
5. Determine the angle phi (θ):
FTy = FTsin(θ)
sin(θ) = FTy / FT = 1.96 N / 18.45 N ≈ 0.106
θ = sin^-1(0.106) ≈ 6.09°
Therefore, the magnitude of FT is approximately 18.45 N and the angle phi it makes with the horizontal is approximately 6.09 degrees.
Your approach to finding the horizontal and vertical components of the tension force (FT) is correct, but there seems to be a calculation error in your work. Let's go through the steps again and see if we can find where the mistake lies.
Step 1: Finding the vertical component (FTy)
You correctly determined that the vertical component of FT is equal to the weight of the ball, which is 1.96 N.
Step 2: Finding the linear velocity (v)
You correctly calculated the linear velocity using the formula v = 2πr, where r is the length of the string. Plugging in the values, we get v = 2 * π * 0.6 = 3.77 m/s.
Step 3: Finding the horizontal component (FTx)
To find FTx, we use the formula Fc = mv^2 / r, where Fc is the centripetal force. Plugging in the values, we get FTx = (0.2 * (3.77)^2) / 0.6 = 3.35 N.
Step 4: Finding the magnitude of FT
To find the magnitude of FT, we use the Pythagorean theorem. FT^2 = FTx^2 + FTy^2. Plugging in the values, we get FT^2 = 3.35^2 + 1.96^2 = 15.1921. Taking the square root gives FT = √15.1921 = 3.9 N (approximately).
So, the magnitude of FT is approximately 3.9 N.
Step 5: Finding the angle (φ)
To find the angle φ, we use the equation FTy = FT * sin(φ). Plugging in the values, we have 1.96 N = 3.9 N * sin(φ). Dividing both sides by 3.9 N gives sin(φ) = 1.96 / 3.9 = 0.5. Taking the inverse sine (sin^-1) of 0.5 gives us φ = 30 degrees.
Therefore, the magnitude of FT is approximately 3.9 N and the angle φ it makes with the horizontal is 30 degrees.
Double-check your calculations to make sure you didn't make any arithmetic errors.