Quiz 7

Physics 6A

2/26/03

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

A cylindrical spool of radius $R = 0.39 m$ and moment of inertia $I = 0.23 kg m^2$ is mounted on a frictionless axis as shown below. A string of negligible mass is wrapped around the cylinder and the string's free end is attached to a block of mass $m = 2 kg$. The block is initially at rest being suspended only by the string. Take the acceleration of gravity to be $g = 9.8 m/s^2$.

(a) (10 points) After the block falls a distance $d = 0.51 m$, what is it's speed?

(b) (5 points) What is the acceleration of the block?

Hint: This is a simplification of this week's homework problem ``Hanging masses on pulley" which is solved using conservation of energy.

Useful information: Kinetic energy of a rotating body: $K = {1\over 2} I \omega^2$. Gravitational potential energy: $U = mgy$. Velocity at rim of wheel $v = \omega R$.

Solution

(a): Speed after falling

Initial energy: Choose $y = 0$ at the initial position of the block. Then there is no potential energy. Since nothing is moving there's no kinetic energy. Therefore the total energy $E = 0$.

Final energy: $U = -mgd = -10.00 J$. Kinetic energy of block $K_b = {1\over 2}m v^2$. Rotational kinetic energy of cylinder $K = {1\over 2}I \omega^2 = {1\over 2}I (v/R)^2$. So the total energy is the sum of these three terms: $E_{total} = -mgd + {1\over 2}(m v^2 + (I/R^2)v^2) = -mgd + {1\over 2}(m + I/R^2)v^2$.

Here $I/R^2 = 1.51kg$

equating initial energy with final energy: $E_{total} = -mgd + {1\over 2}(m + I/R^2)v^2 = 0$. Solving for $v^2$:

$$v^2= {2mgd \over m + I/R^2} = 5.69 m^2/s^2$$ (1)

so $v = 2.39 m/s$.

(b): Acceleration

Note that from above we have

$v^2= 2{mg \over m + I/R^2}d = 2 \; (5.58m/s^2) \;d$

Compare this with the formula for the velocity as a function of position for a 1d system moving with constant acceleration $a$: $v^2-v_0^2 = 2a(x-x_0)$. Here $v_0 = 0$ and $d = x-x_0$. So $v^2 = 2a d$. We see that means that $a = {mg \over m + I/R^2}$ so $a = 5.58m/s^2$.

Quiz 7

Physics 6A

2/26/03

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

A cylindrical spool of radius $R = 0.35 m$ and moment of inertia $I = 0.07 kg m^2$ is mounted on a frictionless axis as shown below. A string of negligible mass is wrapped around the cylinder and the string's free end is attached to a block of mass $m = 1 kg$. The block is initially at rest being suspended only by the string. Take the acceleration of gravity to be $g = 9.8 m/s^2$.

(a) (10 points) After the block falls a distance $d = 0.88 m$, what is it's speed?

(b) (5 points) What is the acceleration of the block?

Hint: This is a simplification of this week's homework problem ``Hanging masses on pulley" which is solved using conservation of energy.

Useful information: Kinetic energy of a rotating body: $K = {1\over 2} I \omega^2$. Gravitational potential energy: $U = mgy$. Velocity at rim of wheel $v = \omega R$.

Solution

(a): Speed after falling

Initial energy: Choose $y = 0$ at the initial position of the block. Then there is no potential energy. Since nothing is moving there's no kinetic energy. Therefore the total energy $E = 0$.

Final energy: $U = -mgd = -8.62 J$. Kinetic energy of block $K_b = {1\over 2}m v^2$. Rotational kinetic energy of cylinder $K = {1\over 2}I \omega^2 = {1\over 2}I (v/R)^2$. So the total energy is the sum of these three terms: $E_{total} = -mgd + {1\over 2}(m v^2 + (I/R^2)v^2) = -mgd + {1\over 2}(m + I/R^2)v^2$.

Here $I/R^2 = 0.57kg$

equating initial energy with final energy: $E_{total} = -mgd + {1\over 2}(m + I/R^2)v^2 = 0$. Solving for $v^2$:

$$v^2= {2mgd \over m + I/R^2} = 10.98 m^2/s^2$$ (2)

so $v = 3.31 m/s$.

(b): Acceleration

Note that from above we have

$v^2= 2{mg \over m + I/R^2}d = 2 \; (6.24m/s^2) \;d$

Compare this with the formula for the velocity as a function of position for a 1d system moving with constant acceleration $a$: $v^2-v_0^2 = 2a(x-x_0)$. Here $v_0 = 0$ and $d = x-x_0$. So $v^2 = 2a d$. We see that means that $a = {mg \over m + I/R^2}$ so $a = 6.24m/s^2$.

Quiz 7

Physics 6A

2/26/03

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

A cylindrical spool of radius $R = 0.31 m$ and moment of inertia $I = 0.14 kg m^2$ is mounted on a frictionless axis as shown below. A string of negligible mass is wrapped around the cylinder and the string's free end is attached to a block of mass $m = 3 kg$. The block is initially at rest being suspended only by the string. Take the acceleration of gravity to be $g = 9.8 m/s^2$.

(a) (10 points) After the block falls a distance $d = 0.45 m$, what is it's speed?

(b) (5 points) What is the acceleration of the block?

Hint: This is a simplification of this week's homework problem ``Hanging masses on pulley" which is solved using conservation of energy.

Useful information: Kinetic energy of a rotating body: $K = {1\over 2} I \omega^2$. Gravitational potential energy: $U = mgy$. Velocity at rim of wheel $v = \omega R$.

Solution

(a): Speed after falling

Initial energy: Choose $y = 0$ at the initial position of the block. Then there is no potential energy. Since nothing is moving there's no kinetic energy. Therefore the total energy $E = 0$.

Final energy: $U = -mgd = -13.23 J$. Kinetic energy of block $K_b = {1\over 2}m v^2$. Rotational kinetic energy of cylinder $K = {1\over 2}I \omega^2 = {1\over 2}I (v/R)^2$. So the total energy is the sum of these three terms: $E_{total} = -mgd + {1\over 2}(m v^2 + (I/R^2)v^2) = -mgd + {1\over 2}(m + I/R^2)v^2$.

Here $I/R^2 = 1.46kg$

equating initial energy with final energy: $E_{total} = -mgd + {1\over 2}(m + I/R^2)v^2 = 0$. Solving for $v^2$:

$$v^2= {2mgd \over m + I/R^2} = 5.94 m^2/s^2$$ (3)

so $v = 2.44 m/s$.

(b): Acceleration

Note that from above we have

$v^2= 2{mg \over m + I/R^2}d = 2 \; (6.60m/s^2) \;d$

Compare this with the formula for the velocity as a function of position for a 1d system moving with constant acceleration $a$: $v^2-v_0^2 = 2a(x-x_0)$. Here $v_0 = 0$ and $d = x-x_0$. So $v^2 = 2a d$. We see that means that $a = {mg \over m + I/R^2}$ so $a = 6.60m/s^2$.

Quiz 7

Physics 6A

2/26/03

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

A cylindrical spool of radius $R = 0.27 m$ and moment of inertia $I = 0.12 kg m^2$ is mounted on a frictionless axis as shown below. A string of negligible mass is wrapped around the cylinder and the string's free end is attached to a block of mass $m = 2 kg$. The block is initially at rest being suspended only by the string. Take the acceleration of gravity to be $g = 9.8 m/s^2$.

(a) (10 points) After the block falls a distance $d = 0.83 m$, what is it's speed?

(b) (5 points) What is the acceleration of the block?

Hint: This is a simplification of this week's homework problem ``Hanging masses on pulley" which is solved using conservation of energy.

Useful information: Kinetic energy of a rotating body: $K = {1\over 2} I \omega^2$. Gravitational potential energy: $U = mgy$. Velocity at rim of wheel $v = \omega R$.

Solution

(a): Speed after falling

Initial energy: Choose $y = 0$ at the initial position of the block. Then there is no potential energy. Since nothing is moving there's no kinetic energy. Therefore the total energy $E = 0$.

Final energy: $U = -mgd = -16.27 J$. Kinetic energy of block $K_b = {1\over 2}m v^2$. Rotational kinetic energy of cylinder $K = {1\over 2}I \omega^2 = {1\over 2}I (v/R)^2$. So the total energy is the sum of these three terms: $E_{total} = -mgd + {1\over 2}(m v^2 + (I/R^2)v^2) = -mgd + {1\over 2}(m + I/R^2)v^2$.

Here $I/R^2 = 1.65kg$

equating initial energy with final energy: $E_{total} = -mgd + {1\over 2}(m + I/R^2)v^2 = 0$. Solving for $v^2$:

$$v^2= {2mgd \over m + I/R^2} = 8.92 m^2/s^2$$ (4)

so $v = 2.99 m/s$.

(b): Acceleration

Note that from above we have

$v^2= 2{mg \over m + I/R^2}d = 2 \; (5.38m/s^2) \;d$

Compare this with the formula for the velocity as a function of position for a 1d system moving with constant acceleration $a$: $v^2-v_0^2 = 2a(x-x_0)$. Here $v_0 = 0$ and $d = x-x_0$. So $v^2 = 2a d$. We see that means that $a = {mg \over m + I/R^2}$ so $a = 5.38m/s^2$.

Quiz 7

Physics 6A

2/26/03

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

A cylindrical spool of radius $R = 0.23 m$ and moment of inertia $I = 0.03 kg m^2$ is mounted on a frictionless axis as shown below. A string of negligible mass is wrapped around the cylinder and the string's free end is attached to a block of mass $m = 1 kg$. The block is initially at rest being suspended only by the string. Take the acceleration of gravity to be $g = 9.8 m/s^2$.

(a) (10 points) After the block falls a distance $d = 0.4 m$, what is it's speed?

(b) (5 points) What is the acceleration of the block?

Hint: This is a simplification of this week's homework problem ``Hanging masses on pulley" which is solved using conservation of energy.

Useful information: Kinetic energy of a rotating body: $K = {1\over 2} I \omega^2$. Gravitational potential energy: $U = mgy$. Velocity at rim of wheel $v = \omega R$.

Solution

(a): Speed after falling

Initial energy: Choose $y = 0$ at the initial position of the block. Then there is no potential energy. Since nothing is moving there's no kinetic energy. Therefore the total energy $E = 0$.

Final energy: $U = -mgd = -3.92 J$. Kinetic energy of block $K_b = {1\over 2}m v^2$. Rotational kinetic energy of cylinder $K = {1\over 2}I \omega^2 = {1\over 2}I (v/R)^2$. So the total energy is the sum of these three terms: $E_{total} = -mgd + {1\over 2}(m v^2 + (I/R^2)v^2) = -mgd + {1\over 2}(m + I/R^2)v^2$.

Here $I/R^2 = 0.57kg$

equating initial energy with final energy: $E_{total} = -mgd + {1\over 2}(m + I/R^2)v^2 = 0$. Solving for $v^2$:

$$v^2= {2mgd \over m + I/R^2} = 5.00 m^2/s^2$$ (5)

so $v = 2.24 m/s$.

(b): Acceleration

Note that from above we have

$v^2= 2{mg \over m + I/R^2}d = 2 \; (6.25m/s^2) \;d$

Compare this with the formula for the velocity as a function of position for a 1d system moving with constant acceleration $a$: $v^2-v_0^2 = 2a(x-x_0)$. Here $v_0 = 0$ and $d = x-x_0$. So $v^2 = 2a d$. We see that means that $a = {mg \over m + I/R^2}$ so $a = 6.25m/s^2$.

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