Question

A 2kg puck slides with uniform circular motion on a smooth surface. When the radius is 0.80m, the tension in the rope connecting the puck to the center of rotation is 15N.

Find:
a.) Free body diagram of the puck (looking at it from the side) when the center of rotation is to the left of it.

b.) The angular velocity of the system

c.) The acceleration of the system

d.) The period of the system.

e.) If the radius is increased to 1.60m but the angular velocity remains constant, determine the tension in the rope

f.) If the radius is 0.80m, but the angular velocity is doubled, determine the tension in the rope.

Answer 1

1. Free Body Diagram (FBD):

- Tension (T) horizontally.

- Weight (mg) vertically.

2. Angular Velocity (
`\(\omega\)`):

-
`\(\omega \approx 2.74 \, \text{rad/s}\).`

3. Acceleration (a):

-
`\(a_c \approx 8.94 \, \text{m/s}^2\).`

4. Period (T):

-
`\(T \approx 2.30 \, \text{s}\).`

5. Tension with Increased Radius (
`\(r = 1.60 \, \text{m}\)`):

-
`\(T' \approx 60 \, \text{N}\).`

6. Tension with Doubled Angular Velocity (
`\(\omega' = 2 \cdot \omega\)`):

-
`\(T'' \approx 60 \, \text{N}\).`

Let's break down the problem step by step:

a.) Free Body Diagram (FBD) of the puck:

The puck is undergoing circular motion, so there must be a centripetal force acting towards the center. The only horizontal force is the tension in the rope. The weight (mg) acts vertically downward.

### b.) Angular velocity (
`\(\omega\)`):

The centripetal force (Fc) is provided by the tension in the rope:

`\[ F_c = T \]`

The centripetal force is also given by the formula
`\(F_c = (mv^2)/(r)\)`, where v is the tangential velocity. In uniform circular motion,
`\(v = r \cdot \omega\)`. Setting these two equations equal to each other:

`\[ T = (m(\omega r)^2)/(r) \]`

Solving for
`\(\omega\):`

`\[ \omega = \sqrt{(T)/(m)} \]`

Substitute the given values:

`\[ \omega = \sqrt{\frac{15 \, \text{N}}{2 \, \text{kg}}} \]`

`\[ \omega \approx 2.74 \, \text{rad/s} \]`

### c.) Acceleration (a):

The centripetal acceleration (
`\(a_c\)`) is given by
`\(a_c = (v^2)/(r)\)`, where v is the tangential velocity. In uniform circular motion,
`\(v = r \cdot \omega\):`

`\[ a_c = ((r \cdot \omega)^2)/(r) \]`

Substitute the values:

`\[ a_c = \frac{(0.80 \, \text{m} \cdot 2.74 \, \text{rad/s})^2}{0.80 \, \text{m}} \]`

`\[ a_c \approx 8.9375 \, \text{m/s}^2 \]`

### d.) Period (T):

The period is the time it takes for one complete revolution, and it's related to angular velocity (
`\(\omega\)`) by the formula
`\(T = (2\pi)/(\omega)\):`

`\[ T = \frac{2\pi}{2.74 \, \text{rad/s}} \]`

`\[ T \approx 2.30 \, \text{s} \]`

e.) Tension with increased radius (
`\(r = 1.60 \, \text{m}\)`):

Using the centripetal force formula (
`\(F_c = (mv^2)/(r)\)`) and the fact that
`\(v = r \cdot \omega\)`, we can write the tension as:

`\[ T = (m(\omega r)^2)/(r) \]`

Since
`\(\omega\)` is constant:

`\[ T' = (m(\omega r')^2)/(r') \]`

`\[ T' = \frac{m(\omega \cdot 1.60 \, \text{m})^2}{1.60 \, \text{m}} \]`

`\[ T' \approx 60 \, \text{N} \]`

f.) Tension with doubled angular velocity (
`\(\omega' = 2 \cdot \omega\)`):

`\[ T'' = (m(\omega' r)^2)/(r) \]`

`\[ T'' = \frac{m((2 \cdot \omega) \cdot 0.80 \, \text{m})^2}{0.80 \, \text{m}} \]`

`\[ T'' \approx 60 \, \text{N} \]`

In both cases (increased radius and doubled angular velocity), the tension in the rope remains approximately
`\(60 \, \text{N}\).`