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2 years ago

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A scientist makes a device to catch baseballs. A long bar of total mass 2.2kg and length 1.2m is fixed at its center. It catches

a ball of mass 0.15kg moving at v=16m/s in a box at one of its ends. Ignore the mass of the catcher box for this problem. What is the angular velocity of the system after the ball is caught?

1 answer:

Valentin [98]2 years ago

8 0

Answer:

ωf = 4.53 rad/s

Explanation:

By conservation of the angular momentum:

Ib*ωb = (Ib + Ic)*ωf

Where

Ib is the inertia of the ball

ωb is the initial angular velocity of the ball

Ic is the inertia of the catcher

ωf is the final angular velocity of the system

We need to calculate first Ib, Ic, ωb:

$Ib = mb*(L/2)^2=0.15*(1.2/2)^2=0.054 kg.m^2$

$Ic = mc/12*L^2=2.2/12*1.2^2=0.264 kg.m^2$

ωb = Vb / (L/2) = 16 / (1.2/2) = 26.67 m/s

Now, ωf will be:

$\omega f = \frac{Ib*\omega b}{Ib + Ic} = 4.53rad/s$

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Starting from rest, a basketball rolls from top of a hill to the bottom, reaching a translational speed of 6.8 m/s. Ignore frict

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Answer:

Explanation:

for baseball

(a) Let the mass of the baseball is m.

radius of baseball is r.

Total kinetic energy of the baseball, T = rotational kinetic energy + translational kinetic energy

T = 0.5 Iω² + 0.5 mv²

Where, I be the moment of inertia and ω be the angular speed.

ω = v/r

T = 0.5 x 2/3 mr² x v²/r² + 0.5 mv²

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According to the conservation of energy, the total kinetic energy at the bottom is equal to the total potential energy at the top.

m g h = 0.83 mv²

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(b) Let the velocity of juice can is v'.

moment of inertia of the juice can = 1/2mr²

So, total kinetic energy

T = 0.5 x I x ω² + 0.5 mv²

T = 0.5 x 0.5 x m x r² x v²/r² + 0.5 mv²

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3 years ago

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

so

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

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