Why does it takes the outer planets so long to orbit the Sun? Try to come up with two reasons.

A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

2 years ago

You are sitting in a car that is travelling at a constant 10 m/s. How fast is your drink, sitting in the cup holder, traveling r

Ksenya-84 [330]

Here, we are required to determine how fast is you drink, sitting in the cup holder, travelling relative to the car.

The speed of the drink, sitting in the cup holder, relative to the car is; 0m/s

From the laws of relative motion,

when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)

when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)when object A and Object B are travelling with speed a and b respectively in opposite directions, the speed of Object A relative to B is; (a+b)

when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)when object A and Object B are travelling with speed a and b respectively in opposite directions, the speed of Object A relative to B is; (a+b)when object A and Object B are travelling with speed a and b respectively in the same direction, where speed a = speed b, then the speed of object A relative to object B is; zero(0).

Evidently, the scenario in the question is similar to the third scenario above. The cup, sitting in the cup holder is travelling with the car at the same constant speed 10m/s.

Therefore, the speed of the drink relative to the car is zero(0).

Read more:

brainly.com/question/20549055

7 0

1 year ago

A large heavy truck smashes into a brick wall totally crumpling the front end. what is the conclusion of this description

Oduvanchick [21]

The person driving the truck was killed
the wall was destroyed

8 0

3 years ago

A motorcycle starting from rest covers 200 metre distance in 6 second. Calculate final velocity and acceleration of the motorcyc

QveST [7]