An object of mass 0.675 kg on a frictionless table is attached to a string that passes through a hole in the table at the center of horizontal circle in which the object moves with constant speed. (a) If the radius of the circle is 0.500 m and the speed is 10.0 m/s, compute the tension in the string. (b) It is found that drawing an additional 0.200 m of the string down through the hole, thereby reducing the radius of the circle to 0.300 m, has the effect of multiplying the original tension in the string by 4.63. Compute the total work done by the string on the revolving object during the reduction of the radius.

To tackle this problem, we need to break it down into two parts: first, calculating the tension in the string when the object is moving in a circle of radius 0.500 m at a speed of 10.0 m/s, and second, determining the work done by the string when the radius is reduced to 0.300 m, which increases the tension significantly. Let's dive into each part step by step.

Part (a): Calculating Tension in the String

The tension in the string provides the necessary centripetal force to keep the object moving in a circular path. The formula for centripetal force (Fc) is given by:

• Fc = (m * v²) / r

Where:

• m = mass of the object (0.675 kg)
• v = speed of the object (10.0 m/s)
• r = radius of the circle (0.500 m)

Now, substituting the values into the formula:

• Fc = (0.675 kg * (10.0 m/s)²) / 0.500 m
• Fc = (0.675 kg * 100 m²/s²) / 0.500 m
• Fc = 67.5 kg·m/s² / 0.500 m
• Fc = 135 N

Thus, the tension in the string when the radius is 0.500 m is 135 N.

Part (b): Work Done During Radius Reduction

Next, we need to calculate the work done by the string when the radius is reduced from 0.500 m to 0.300 m, which results in the tension increasing by a factor of 4.63. First, we find the new tension:

• New Tension (T') = 4.63 * Original Tension
• T' = 4.63 * 135 N = 624.55 N

Now, to find the work done (W) by the tension as the radius changes, we can use the formula:

• W = T * d

Where:

• T = average tension during the process
• d = distance over which the tension acts (0.200 m)

To find the average tension, we can take the average of the initial and final tensions:

• Average Tension = (Original Tension + New Tension) / 2
• Average Tension = (135 N + 624.55 N) / 2 = 379.775 N

Now, substituting this average tension into the work formula:

• W = 379.775 N * 0.200 m
• W = 75.955 N·m

Thus, the total work done by the string on the revolving object during the reduction of the radius is approximately 75.96 J.

Summary

In summary, we found that the tension in the string when the radius is 0.500 m is 135 N. After reducing the radius to 0.300 m, the tension increased to 624.55 N, and the work done by the string during this process was approximately 75.96 J. This illustrates how changes in radius affect both tension and the work done in circular motion.