A 0.5 kg mass is moving in a circle of radius 1.8 m on a flat frictionless table at the end of a string. The speed of the mass is 2.6 m/s. The string routes through a hole in the center of the table and is held by you underneath the table.

A.) What is the angular momentum of the mass?
B.) What is the tension in the string?
C.) If you pull on the string so that the radius of the circle decreases to one half its former value, what is the new tension in the string?
D.) How much work did you do to reduce the radius by a factor of one-half?

A.) To calculate the angular momentum of the mass, we use the formula:

Angular momentum (L) = mvr

where:
m = mass of the object (0.5 kg)
v = velocity of the object (2.6 m/s)
r = radius of the circle (1.8 m)

Substituting the given values into the formula, we get:

L = (0.5 kg)(2.6 m/s)(1.8 m)
L = 2.34 kg·m²/s

Therefore, the angular momentum of the mass is 2.34 kg·m²/s.

B.) To find the tension in the string, we need to consider the forces acting on the mass in circular motion. In this case, the tension in the string provides the centripetal force.

The centripetal force (F) is given by the formula:

F = mv² / r

where:
m = mass of the object (0.5 kg)
v = velocity of the object (2.6 m/s)
r = radius of the circle (1.8 m)

Substituting the given values into the formula, we get:

F = (0.5 kg)(2.6 m/s)² / 1.8 m
F = 1.69 N

Therefore, the tension in the string is 1.69 N.

C.) If the radius of the circle decreases to half its former value (0.9 m), we can calculate the new tension using the same formula:

F = mv² / r

where:
m = mass of the object (0.5 kg)
v = velocity of the object (2.6 m/s)
r = new radius of the circle (0.9 m)

Substituting the given values into the formula, we get:

F = (0.5 kg)(2.6 m/s)² / 0.9 m
F = 3.01 N

Therefore, the new tension in the string is 3.01 N.

D.) To determine the work done to reduce the radius by a factor of one-half, we need to consider the change in kinetic energy. The work done (W) is given by the formula:

W = ΔKE = KE final - KE initial

Since no external forces are acting on the system, the total mechanical energy is conserved. Therefore, the initial kinetic energy is equal to the final kinetic energy.

The kinetic energy (KE) of an object is given by the formula:

KE = 0.5 mv²

Substituting the given values into the formula, we get:

KE initial = 0.5 (0.5 kg) (2.6 m/s)² = 1.69 J

To calculate the final kinetic energy, we need to find the velocity when the radius is reduced. Since angular momentum is conserved in this type of motion, we can use the formula:

L = mvr

Rearranging the formula to solve for velocity (v), we get:

v = L / (mr)

Substituting the given values into the formula, we get:

v = (2.34 kg·m²/s) / ((0.5 kg)(0.9 m))
v = 10.44 m/s

Now, we can find the final kinetic energy using the formula for kinetic energy:

KE final = 0.5 (0.5 kg) (10.44 m/s)² = 28.7 J

Finally, we can calculate the work done using the formula for the change in kinetic energy:

W = KE final - KE initial = 28.7 J - 1.69 J = 27.01 J

Therefore, the work done to reduce the radius by a factor of one-half is 27.01 J.

To calculate the answers, we need to use some basic equations related to circular motion. Let's go through each question step by step:

A.) What is the angular momentum of the mass?

Angular momentum (L) is defined as the product of moment of inertia (I) and angular velocity (ω). For a point mass moving in a circular path, the moment of inertia is simply the mass (m) multiplied by the square of the radius (r squared). The angular velocity (ω) is the speed (v) divided by the radius (r).

So, we can determine angular momentum using the formula:
L = I * ω = m * r^2 * ω = m * r * v

Given:
m = 0.5 kg (mass)
r = 1.8 m (radius)
v = 2.6 m/s (speed)

Substituting these values into the formula:
L = 0.5 kg * 1.8 m * 2.6 m/s = 2.34 kg·m^2/s

Therefore, the angular momentum of the mass is 2.34 kg·m^2/s.

B.) What is the tension in the string?

The tension in the string (T) provides the centripetal force required to keep the mass moving in a circular path.

Using the centripetal force formula:
F = m * ω^2 * r

We know that the centripetal force is provided by the tension in the string, so we can equate the two:
T = m * ω^2 * r

Substituting the given values:
T = 0.5 kg * (2.6 m/s / 1.8 m)^2 * 1.8 m = 4.17 N

Therefore, the tension in the string is 4.17 N.

C.) If you pull on the string so that the radius of the circle decreases to one half its former value, what is the new tension in the string?

When the radius decreases to one half its former value, the new radius (r') is 1.8 m / 2 = 0.9 m.

Using the same formula as in part B, we can calculate the new tension (T') with the updated radius:
T' = m * ω^2 * r' = 0.5 kg * (2.6 m/s / 0.9 m)^2 * 0.9 m = 14.07 N

Therefore, the new tension in the string is 14.07 N.

D.) How much work did you do to reduce the radius by a factor of one-half?

The work done (W) can be calculated using the equation:
W = ΔKE = KE_final - KE_initial

Since the mass is moving in a circular path and there is no change in its speed, there is no change in its kinetic energy (KE) or work done.

Therefore, the work done to reduce the radius by a factor of one-half is zero.

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