QuestionLearning Goal: use static equilibrium principles, Newton's second law for rotation, and center-of-mass energy to analyze a suspended meterstick system.
A meterstick ( $L=1.00 \mathrm{~m}$ ) has a uniformly distributed mass of $m=0.210 \mathrm{~kg}$ . Initially, it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.
After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction. ${ }^{\infty}$ What is the magnitude $T_{\mathrm{v}}^{\prime}$ of the tension in the string when the meterstick is vertical? $T_{\mathrm{v}}^{\prime}=$ $\square$ N

Studdy Solution

STEP 1

What is this asking? How much tension is in the string when the meterstick is hanging straight down after we cut the other string? Watch out! Don't forget that the meterstick rotates around the point where the string is attached, not its center of mass!

STEP 2

1. Find the change in height of the center of mass.
2. Calculate the change in potential energy.
3. Calculate the rotational kinetic energy.
4. Find the angular velocity.
5. Calculate the centripetal acceleration.
6. Determine the tension in the string.

STEP 3

Initially, the meterstick's center of mass is at the $\frac{L}{2} = \frac{1.00 \text{ m}}{2} = 0.50 \text{ m}$ mark.

STEP 4

The string is attached at the $0.25 \text{ m}$ mark.

STEP 5

When the meterstick is vertical, the center of mass has fallen by a distance of $0.50 \text{ m} - 0.25 \text{ m} = 0.25 \text{ m}$ .
This change in height is $\Delta h = 0.25 \text{ m}$ .

STEP 6

The change in potential energy is given by $\Delta U = m \cdot g \cdot \Delta h$ , where $m = 0.210 \text{ kg}$ is the mass, $g = 9.8 \frac{\text{m}}{\text{s}^2}$ is the acceleration due to gravity, and $\Delta h = 0.25 \text{ m}$ is the change in height.

STEP 7

So, $\Delta U = 0.210 \text{ kg} \cdot 9.8 \frac{\text{m}}{\text{s}^2} \cdot 0.25 \text{ m} = 0.5145 \text{ J}$ .

STEP 8

By conservation of energy, the change in potential energy is converted into rotational kinetic energy, $K = \frac{1}{2} I \omega^2$ , where $I$ is the moment of inertia and $\omega$ is the angular velocity.

STEP 9

Thus, $\Delta U = K$ , so $0.5145 \text{ J} = \frac{1}{2} I \omega^2$ .

STEP 10

The moment of inertia of a rod rotating about one end is $I = \frac{1}{3} m L^2$ .
Here, $I = \frac{1}{3} \cdot 0.210 \text{ kg} \cdot (1.00 \text{ m})^2 = 0.07 \text{ kg} \cdot \text{m}^2$ .

STEP 11

Substituting this into the equation from the previous step, we get $0.5145 \text{ J} = \frac{1}{2} (0.07 \text{ kg} \cdot \text{m}^2) \omega^2$ .

STEP 12

Solving for $\omega$ , we find $\omega^2 = \frac{2 \cdot 0.5145 \text{ J}}{0.07 \text{ kg} \cdot \text{m}^2} = 14.7 \frac{1}{\text{s}^2}$ , so $\omega = \sqrt{14.7 \frac{1}{\text{s}^2}} \approx 3.834 \frac{1}{\text{s}}$ .

STEP 13

The centripetal acceleration of the center of mass is $a_c = \omega^2 r$ , where $r$ is the distance from the pivot point to the center of mass, which is $0.25 \text{ m}$ .

STEP 14

So, $a_c = (14.7 \frac{1}{\text{s}^2}) \cdot 0.25 \text{ m} = 3.675 \frac{\text{m}}{\text{s}^2}$ .

STEP 15

When the meterstick is vertical, the net force on the center of mass is the tension $T$ upwards minus the weight $mg$ downwards, which equals the mass times the centripetal acceleration: $T - mg = ma_c$ .

STEP 16

Therefore, $T = mg + ma_c = m(g + a_c) = 0.210 \text{ kg} (9.8 \frac{\text{m}}{\text{s}^2} + 3.675 \frac{\text{m}}{\text{s}^2}) = 0.210 \text{ kg} \cdot 13.475 \frac{\text{m}}{\text{s}^2} \approx 2.83 \text{ N}$ .

STEP 17

The tension in the string when the meterstick is vertical is $2.83 \text{ N}$ .

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Studdy Solution

STEP 1

What is this asking? How much tension is in the string when the meterstick is hanging straight down after we cut the other string? Watch out! Don't forget that the meterstick rotates around the point where the string is attached, not its center of mass!

STEP 2

1. Find the change in height of the center of mass.
2. Calculate the change in potential energy.
3. Calculate the rotational kinetic energy.
4. Find the angular velocity.
5. Calculate the centripetal acceleration.
6. Determine the tension in the string.

STEP 3

Initially, the meterstick's center of mass is at the $\frac{L}{2} = \frac{1.00 \text{ m}}{2} = 0.50 \text{ m}$ mark.

STEP 4

The string is attached at the $0.25 \text{ m}$ mark.

STEP 5

When the meterstick is vertical, the center of mass has fallen by a distance of $0.50 \text{ m} - 0.25 \text{ m} = 0.25 \text{ m}$ .
This change in height is $\Delta h = 0.25 \text{ m}$ .

STEP 6

The change in potential energy is given by $\Delta U = m \cdot g \cdot \Delta h$ , where $m = 0.210 \text{ kg}$ is the mass, $g = 9.8 \frac{\text{m}}{\text{s}^2}$ is the acceleration due to gravity, and $\Delta h = 0.25 \text{ m}$ is the change in height.

STEP 7

So, $\Delta U = 0.210 \text{ kg} \cdot 9.8 \frac{\text{m}}{\text{s}^2} \cdot 0.25 \text{ m} = 0.5145 \text{ J}$ .

STEP 8

By conservation of energy, the change in potential energy is converted into rotational kinetic energy, $K = \frac{1}{2} I \omega^2$ , where $I$ is the moment of inertia and $\omega$ is the angular velocity.

STEP 9

Thus, $\Delta U = K$ , so $0.5145 \text{ J} = \frac{1}{2} I \omega^2$ .

STEP 10

The moment of inertia of a rod rotating about one end is $I = \frac{1}{3} m L^2$ .
Here, $I = \frac{1}{3} \cdot 0.210 \text{ kg} \cdot (1.00 \text{ m})^2 = 0.07 \text{ kg} \cdot \text{m}^2$ .

STEP 11

Substituting this into the equation from the previous step, we get $0.5145 \text{ J} = \frac{1}{2} (0.07 \text{ kg} \cdot \text{m}^2) \omega^2$ .

STEP 12

Solving for $\omega$ , we find $\omega^2 = \frac{2 \cdot 0.5145 \text{ J}}{0.07 \text{ kg} \cdot \text{m}^2} = 14.7 \frac{1}{\text{s}^2}$ , so $\omega = \sqrt{14.7 \frac{1}{\text{s}^2}} \approx 3.834 \frac{1}{\text{s}}$ .

STEP 13

The centripetal acceleration of the center of mass is $a_c = \omega^2 r$ , where $r$ is the distance from the pivot point to the center of mass, which is $0.25 \text{ m}$ .

STEP 14

So, $a_c = (14.7 \frac{1}{\text{s}^2}) \cdot 0.25 \text{ m} = 3.675 \frac{\text{m}}{\text{s}^2}$ .

STEP 15

When the meterstick is vertical, the net force on the center of mass is the tension $T$ upwards minus the weight $mg$ downwards, which equals the mass times the centripetal acceleration: $T - mg = ma_c$ .

STEP 16

Therefore, $T = mg + ma_c = m(g + a_c) = 0.210 \text{ kg} (9.8 \frac{\text{m}}{\text{s}^2} + 3.675 \frac{\text{m}}{\text{s}^2}) = 0.210 \text{ kg} \cdot 13.475 \frac{\text{m}}{\text{s}^2} \approx 2.83 \text{ N}$ .

STEP 17

The tension in the string when the meterstick is vertical is $2.83 \text{ N}$ .

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Studdy Solution

STEP 1

What is this asking? How much tension is in the string when the meterstick is hanging straight down after we cut the other string? Watch out! Don't forget that the meterstick rotates around the point where the string is attached, not its center of mass!

STEP 2

1. Find the change in height of the center of mass.2. Calculate the change in potential energy.3. Calculate the rotational kinetic energy.4. Find the angular velocity.5. Calculate the centripetal acceleration.6. Determine the tension in the string.

STEP 3

Initially, the meterstick's center of mass is at the L2=1.00 m2=0.50 m\frac{L}{2} = \frac{1.00 \text{ m}}{2} = 0.50 \text{ m}2L​=21.00 m​=0.50 m mark.

STEP 4

The string is attached at the 0.25 m0.25 \text{ m}0.25 m mark.

STEP 5

When the meterstick is vertical, the center of mass has fallen by a distance of 0.50 m−0.25 m=0.25 m0.50 \text{ m} - 0.25 \text{ m} = 0.25 \text{ m}0.50 m−0.25 m=0.25 m.This **change in height** is Δh=0.25 m\Delta h = 0.25 \text{ m}Δh=0.25 m.

STEP 6

STEP 7

So, ΔU=0.210 kg⋅9.8ms2⋅0.25 m=0.5145 J\Delta U = 0.210 \text{ kg} \cdot 9.8 \frac{\text{m}}{\text{s}^2} \cdot 0.25 \text{ m} = 0.5145 \text{ J}ΔU=0.210 kg⋅9.8s2m​⋅0.25 m=0.5145 J.

STEP 8

By conservation of energy, the **change in potential energy** is converted into **rotational kinetic energy**, K=12Iω2K = \frac{1}{2} I \omega^2K=21​Iω2, where III is the moment of inertia and ω\omegaω is the angular velocity.

STEP 9

Thus, ΔU=K\Delta U = KΔU=K, so 0.5145 J=12Iω20.5145 \text{ J} = \frac{1}{2} I \omega^20.5145 J=21​Iω2.

STEP 10

The **moment of inertia** of a rod rotating about one end is I=13mL2I = \frac{1}{3} m L^2I=31​mL2.Here, I=13⋅0.210 kg⋅(1.00 m)2=0.07 kg⋅m2I = \frac{1}{3} \cdot 0.210 \text{ kg} \cdot (1.00 \text{ m})^2 = 0.07 \text{ kg} \cdot \text{m}^2I=31​⋅0.210 kg⋅(1.00 m)2=0.07 kg⋅m2.

STEP 11

Substituting this into the equation from the previous step, we get 0.5145 J=12(0.07 kg⋅m2)ω20.5145 \text{ J} = \frac{1}{2} (0.07 \text{ kg} \cdot \text{m}^2) \omega^20.5145 J=21​(0.07 kg⋅m2)ω2.

STEP 12

STEP 13

The **centripetal acceleration** of the center of mass is ac=ω2ra_c = \omega^2 rac​=ω2r, where rrr is the distance from the pivot point to the center of mass, which is 0.25 m0.25 \text{ m}0.25 m.

STEP 14

So, ac=(14.71s2)⋅0.25 m=3.675ms2a_c = (14.7 \frac{1}{\text{s}^2}) \cdot 0.25 \text{ m} = 3.675 \frac{\text{m}}{\text{s}^2}ac​=(14.7s21​)⋅0.25 m=3.675s2m​.

STEP 15

When the meterstick is vertical, the **net force** on the center of mass is the **tension** TTT upwards minus the **weight** mgmgmg downwards, which equals the **mass** times the **centripetal acceleration**: T−mg=macT - mg = ma_cT−mg=mac​.

STEP 16

STEP 17

The tension in the string when the meterstick is vertical is 2.83 N2.83 \text{ N}2.83 N.

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1. Find the change in height of the center of mass.
2. Calculate the change in potential energy.
3. Calculate the rotational kinetic energy.
4. Find the angular velocity.
5. Calculate the centripetal acceleration.
6. Determine the tension in the string.

Initially, the meterstick's center of mass is at the $\frac{L}{2} = \frac{1.00 \text{ m}}{2} = 0.50 \text{ m}$ mark.

The string is attached at the $0.25 \text{ m}$ mark.

When the meterstick is vertical, the center of mass has fallen by a distance of $0.50 \text{ m} - 0.25 \text{ m} = 0.25 \text{ m}$ .
This change in height is $\Delta h = 0.25 \text{ m}$ .

So, $\Delta U = 0.210 \text{ kg} \cdot 9.8 \frac{\text{m}}{\text{s}^2} \cdot 0.25 \text{ m} = 0.5145 \text{ J}$ .

By conservation of energy, the change in potential energy is converted into rotational kinetic energy, $K = \frac{1}{2} I \omega^2$ , where $I$ is the moment of inertia and $\omega$ is the angular velocity.

Thus, $\Delta U = K$ , so $0.5145 \text{ J} = \frac{1}{2} I \omega^2$ .

The moment of inertia of a rod rotating about one end is $I = \frac{1}{3} m L^2$ .
Here, $I = \frac{1}{3} \cdot 0.210 \text{ kg} \cdot (1.00 \text{ m})^2 = 0.07 \text{ kg} \cdot \text{m}^2$ .

Substituting this into the equation from the previous step, we get $0.5145 \text{ J} = \frac{1}{2} (0.07 \text{ kg} \cdot \text{m}^2) \omega^2$ .

The centripetal acceleration of the center of mass is $a_c = \omega^2 r$ , where $r$ is the distance from the pivot point to the center of mass, which is $0.25 \text{ m}$ .

So, $a_c = (14.7 \frac{1}{\text{s}^2}) \cdot 0.25 \text{ m} = 3.675 \frac{\text{m}}{\text{s}^2}$ .

When the meterstick is vertical, the net force on the center of mass is the tension $T$ upwards minus the weight $mg$ downwards, which equals the mass times the centripetal acceleration: $T - mg = ma_c$ .

The tension in the string when the meterstick is vertical is $2.83 \text{ N}$ .