Question41. (II) On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every 2.5 s . If we assume their arms are each 0.80 m long and their individual masses are 55.0 kg , how hard are they pulling on one another?

Studdy Solution

STEP 1

1. The skaters are spinning in a circle with a period of $2.5$ seconds.
2. Each skater has a mass of $55.0$ kg.
3. The distance from the center of the circle to each skater is $0.80$ m.
4. The motion is uniform circular motion.

STEP 2

1. Determine the radius of the circle.
2. Calculate the angular velocity.
3. Determine the centripetal force required for circular motion.
4. Calculate the force exerted by each skater.

STEP 3

Determine the radius of the circle. Since the skaters are holding hands and spinning around a mutual center, the radius $r$ is the length of one arm:
$r = 0.80 \, \text{m}$

STEP 4

Calculate the angular velocity $\omega$ . The formula for angular velocity in terms of the period $T$ is:
$\omega = \frac{2\pi}{T}$
Substitute $T = 2.5 \, \text{s}$ :
$\omega = \frac{2\pi}{2.5} \, \text{rad/s}$

STEP 5

Determine the centripetal force $F_c$ required for circular motion. The formula for centripetal force is:
$F_c = m \cdot r \cdot \omega^2$
Substitute the known values:
$F_c = 55.0 \, \text{kg} \times 0.80 \, \text{m} \times \left(\frac{2\pi}{2.5}\right)^2$

STEP 6

Calculate the force exerted by each skater. Simplify and compute the expression for $F_c$ :
$F_c = 55.0 \times 0.80 \times \left(\frac{2\pi}{2.5}\right)^2$
$F_c \approx 55.0 \times 0.80 \times 6.32$
$F_c \approx 278.3 \, \text{N}$
Thus, each skater is pulling on the other with a force of approximately:
$\boxed{278.3 \, \text{N}}$

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Question41. (II) On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every 2.5 s . If we assume their arms are each 0.80 m long and their individual masses are 55.0 kg , how hard are they pulling on one another?

Studdy Solution

STEP 1

1. The skaters are spinning in a circle with a period of $2.5$ seconds.
2. Each skater has a mass of $55.0$ kg.
3. The distance from the center of the circle to each skater is $0.80$ m.
4. The motion is uniform circular motion.

STEP 2

1. Determine the radius of the circle.
2. Calculate the angular velocity.
3. Determine the centripetal force required for circular motion.
4. Calculate the force exerted by each skater.

STEP 3

Determine the radius of the circle. Since the skaters are holding hands and spinning around a mutual center, the radius $r$ is the length of one arm:
$r = 0.80 \, \text{m}$

STEP 4

Calculate the angular velocity $\omega$ . The formula for angular velocity in terms of the period $T$ is:
$\omega = \frac{2\pi}{T}$
Substitute $T = 2.5 \, \text{s}$ :
$\omega = \frac{2\pi}{2.5} \, \text{rad/s}$

STEP 5

Determine the centripetal force $F_c$ required for circular motion. The formula for centripetal force is:
$F_c = m \cdot r \cdot \omega^2$
Substitute the known values:
$F_c = 55.0 \, \text{kg} \times 0.80 \, \text{m} \times \left(\frac{2\pi}{2.5}\right)^2$

STEP 6

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Question41. (II) On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every 2.5 s . If we assume their arms are each 0.80 m long and their individual masses are 55.0 kg , how hard are they pulling on one another?

Studdy Solution

STEP 1

1. The skaters are spinning in a circle with a period of 2.5 2.5 2.5 seconds.2. Each skater has a mass of 55.0 55.0 55.0 kg.3. The distance from the center of the circle to each skater is 0.80 0.80 0.80 m.4. The motion is uniform circular motion.

STEP 2

1. Determine the radius of the circle.2. Calculate the angular velocity.3. Determine the centripetal force required for circular motion.4. Calculate the force exerted by each skater.

STEP 3

Determine the radius of the circle. Since the skaters are holding hands and spinning around a mutual center, the radius r r r is the length of one arm:r=0.80 m r = 0.80 \, \text{m} r=0.80m

STEP 4

Calculate the angular velocity ω \omega ω. The formula for angular velocity in terms of the period T T T is:ω=2πT \omega = \frac{2\pi}{T} ω=T2π​Substitute T=2.5 s T = 2.5 \, \text{s} T=2.5s:ω=2π2.5 rad/s \omega = \frac{2\pi}{2.5} \, \text{rad/s} ω=2.52π​rad/s

STEP 5

STEP 6

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1. The skaters are spinning in a circle with a period of $2.5$ seconds.
2. Each skater has a mass of $55.0$ kg.
3. The distance from the center of the circle to each skater is $0.80$ m.
4. The motion is uniform circular motion.

1. Determine the radius of the circle.
2. Calculate the angular velocity.
3. Determine the centripetal force required for circular motion.
4. Calculate the force exerted by each skater.

Determine the radius of the circle. Since the skaters are holding hands and spinning around a mutual center, the radius $r$ is the length of one arm:
$r = 0.80 \, \text{m}$

Calculate the angular velocity $\omega$ . The formula for angular velocity in terms of the period $T$ is:
$\omega = \frac{2\pi}{T}$
Substitute $T = 2.5 \, \text{s}$ :
$\omega = \frac{2\pi}{2.5} \, \text{rad/s}$