Quiz 5

Physics 6A

2/12/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

A pear of mass $0.1 kg$ is tied to a string with length $L = 3 m$. The other end of the string is tied to a rigid support. The string is pulled taut and the pear initially makes an angle of $\theta = 25$ degrees with the vertical. It is then released. Take the acceleration of gravity to be $9.8 m/s^2$. Neglect any frictional forces and assume that the size of the pear and the mass of the string are both negligible.

(a) (7 points) What is the speed of the pear at its lowest point of motion?

(b) (8 points) What is the tension in the string at this point?

Solution

(a): Speed at lowest point

Conservation of energy. Choose the coordinate system from the top of the string. The initial height $= -L \cos(\theta) = -2.72 m$.

The final height is $y_f = -L = -3 m$. The initial kinetic energy $K_i = 0$.

Using conservation of energy, we say that the inital energy equals the final energy or $m g y_i = m g y_f + {1\over 2}mv^2$, or $mv^2 = 2 m g(y_i -y_f) = 2 m g L(1-\cos(\theta))$.

So $v = \sqrt{ 2 g L (1-\cos(\theta)} = 2.35m/s$.

(b) Get the tension

F=ma in component form

The net force on the pear is from the tension $T$ and the force of gravity. So $T - m g = m v^2/L$ or $T = mg + m v^2/L$. But from above we know the formula for $mv^2/L = 2 m g (1-\cos(\theta)$ so the tension is

$$T = mg + 2 m g (1-\cos(\theta) = m g (3 - 2\cos(\theta)) = 1.16N$$ (1)

Intermediate variables that might have been used:

$U_i = m g L \cos(\theta) = -2.66J$

$U_f = -m g L = -2.94J$

$\Delta U = U_f - U_i = -0.28J$

$\Delta y = 0.28m$

Quiz 5

Physics 6A

2/12/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

A tangerine of mass $0.17 kg$ is tied to a string with length $L = 3 m$. The other end of the string is tied to a rigid support. The string is pulled taut and the tangerine initially makes an angle of $\theta = 20$ degrees with the vertical. It is then released. Take the acceleration of gravity to be $9.8 m/s^2$. Neglect any frictional forces and assume that the size of the tangerine and the mass of the string are both negligible.

(a) (7 points) What is the speed of the tangerine at its lowest point of motion?

(b) (8 points) What is the tension in the string at this point?

Solution

(a): Speed at lowest point

Conservation of energy. Choose the coordinate system from the top of the string. The initial height $= -L \cos(\theta) = -2.82 m$.

The final height is $y_f = -L = -3 m$. The initial kinetic energy $K_i = 0$.

Using conservation of energy, we say that the inital energy equals the final energy or $m g y_i = m g y_f + {1\over 2}mv^2$, or $mv^2 = 2 m g(y_i -y_f) = 2 m g L(1-\cos(\theta))$.

So $v = \sqrt{ 2 g L (1-\cos(\theta)} = 1.88m/s$.

(b) Get the tension

F=ma in component form

The net force on the tangerine is from the tension $T$ and the force of gravity. So $T - m g = m v^2/L$ or $T = mg + m v^2/L$. But from above we know the formula for $mv^2/L = 2 m g (1-\cos(\theta)$ so the tension is

$$T = mg + 2 m g (1-\cos(\theta) = m g (3 - 2\cos(\theta)) = 1.87N$$ (2)

Intermediate variables that might have been used:

$U_i = m g L \cos(\theta) = -4.70J$

$U_f = -m g L = -5.00J$

$\Delta U = U_f - U_i = -0.30J$

$\Delta y = 0.18m$

Quiz 5

Physics 6A

2/12/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

A grapefruit of mass $0.14 kg$ is tied to a string with length $L = 2 m$. The other end of the string is tied to a rigid support. The string is pulled taut and the grapefruit initially makes an angle of $\theta = 35$ degrees with the vertical. It is then released. Take the acceleration of gravity to be $9.8 m/s^2$. Neglect any frictional forces and assume that the size of the grapefruit and the mass of the string are both negligible.

(a) (7 points) What is the speed of the grapefruit at its lowest point of motion?

(b) (8 points) What is the tension in the string at this point?

Solution

(a): Speed at lowest point

Conservation of energy. Choose the coordinate system from the top of the string. The initial height $= -L \cos(\theta) = -1.64 m$.

The final height is $y_f = -L = -2 m$. The initial kinetic energy $K_i = 0$.

Using conservation of energy, we say that the inital energy equals the final energy or $m g y_i = m g y_f + {1\over 2}mv^2$, or $mv^2 = 2 m g(y_i -y_f) = 2 m g L(1-\cos(\theta))$.

So $v = \sqrt{ 2 g L (1-\cos(\theta)} = 2.66m/s$.

(b) Get the tension

F=ma in component form

The net force on the grapefruit is from the tension $T$ and the force of gravity. So $T - m g = m v^2/L$ or $T = mg + m v^2/L$. But from above we know the formula for $mv^2/L = 2 m g (1-\cos(\theta)$ so the tension is

$$T = mg + 2 m g (1-\cos(\theta) = m g (3 - 2\cos(\theta)) = 1.87N$$ (3)

Intermediate variables that might have been used:

$U_i = m g L \cos(\theta) = -2.25J$

$U_f = -m g L = -2.74J$

$\Delta U = U_f - U_i = -0.50J$

$\Delta y = 0.36m$

Quiz 5

Physics 6A

2/12/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

A tangerine of mass $0.11 kg$ is tied to a string with length $L = 2 m$. The other end of the string is tied to a rigid support. The string is pulled taut and the tangerine initially makes an angle of $\theta = 29$ degrees with the vertical. It is then released. Take the acceleration of gravity to be $9.8 m/s^2$. Neglect any frictional forces and assume that the size of the tangerine and the mass of the string are both negligible.

(a) (7 points) What is the speed of the tangerine at its lowest point of motion?

(b) (8 points) What is the tension in the string at this point?

Solution

(a): Speed at lowest point

Conservation of energy. Choose the coordinate system from the top of the string. The initial height $= -L \cos(\theta) = -1.75 m$.

The final height is $y_f = -L = -2 m$. The initial kinetic energy $K_i = 0$.

Using conservation of energy, we say that the inital energy equals the final energy or $m g y_i = m g y_f + {1\over 2}mv^2$, or $mv^2 = 2 m g(y_i -y_f) = 2 m g L(1-\cos(\theta))$.

So $v = \sqrt{ 2 g L (1-\cos(\theta)} = 2.22m/s$.

(b) Get the tension

F=ma in component form

The net force on the tangerine is from the tension $T$ and the force of gravity. So $T - m g = m v^2/L$ or $T = mg + m v^2/L$. But from above we know the formula for $mv^2/L = 2 m g (1-\cos(\theta)$ so the tension is

$$T = mg + 2 m g (1-\cos(\theta) = m g (3 - 2\cos(\theta)) = 1.35N$$ (4)

Intermediate variables that might have been used:

$U_i = m g L \cos(\theta) = -1.89J$

$U_f = -m g L = -2.16J$

$\Delta U = U_f - U_i = -0.27J$

$\Delta y = 0.25m$

Quiz 5

Physics 6A

2/12/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

A grapefruit of mass $0.18 kg$ is tied to a string with length $L = 1 m$. The other end of the string is tied to a rigid support. The string is pulled taut and the grapefruit initially makes an angle of $\theta = 24$ degrees with the vertical. It is then released. Take the acceleration of gravity to be $9.8 m/s^2$. Neglect any frictional forces and assume that the size of the grapefruit and the mass of the string are both negligible.

(a) (7 points) What is the speed of the grapefruit at its lowest point of motion?

(b) (8 points) What is the tension in the string at this point?

Solution

(a): Speed at lowest point

Conservation of energy. Choose the coordinate system from the top of the string. The initial height $= -L \cos(\theta) = -0.91 m$.

The final height is $y_f = -L = -1 m$. The initial kinetic energy $K_i = 0$.

Using conservation of energy, we say that the inital energy equals the final energy or $m g y_i = m g y_f + {1\over 2}mv^2$, or $mv^2 = 2 m g(y_i -y_f) = 2 m g L(1-\cos(\theta))$.

So $v = \sqrt{ 2 g L (1-\cos(\theta)} = 1.30m/s$.

(b) Get the tension

F=ma in component form

The net force on the grapefruit is from the tension $T$ and the force of gravity. So $T - m g = m v^2/L$ or $T = mg + m v^2/L$. But from above we know the formula for $mv^2/L = 2 m g (1-\cos(\theta)$ so the tension is

$$T = mg + 2 m g (1-\cos(\theta) = m g (3 - 2\cos(\theta)) = 2.07N$$ (5)

Intermediate variables that might have been used:

$U_i = m g L \cos(\theta) = -1.61J$

$U_f = -m g L = -1.76J$

$\Delta U = U_f - U_i = -0.15J$

$\Delta y = 0.09m$

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