Answer:
0.893 rad/s in the clockwise direction
Explanation:
From the law of conservation of angular momentum,
angular momentum before impact = angular momentum after impact
L₁ = L₂
L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)
L₂ = angular momentum of cylinder and angular momentum of bullet after collision.
L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision
So,
L₁ = L₂
L₁ = (I₁ + I₂)ω
ω = L₁/(I₁ + I₂)
ω = L₁/(1/2MR² + mR²)
ω = L₁/(1/2M + m)R²
substituting the values of the variables into the equation, we have
ω = L₁/(1/2M + m)R²
ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²
ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)
ω = + 9 kgm²/s/(2.52 kg)(4 m²)
ω = +9 kgm²/s/10.08 kgm²
ω = + 0.893 rad/s
The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.
The appropriate response is the revolution of the Earth around the Sun. For instance, the Earth finishes one turn about its hub about at regular intervals, yet it finishes one revolution around the Sun about each 365 days. Anyway, the fundamental motivation behind why the planets spin around, or circle, the Sun, is that the gravity of the Sun keeps them in their circles.
Answer:
15.68 m/s
Explanation:
Given that,
She catches the ball 3.2 s later at the same height from which it was thrown.
When it reaches the maximum height, its height is equal to 0.
It will move under the action of gravity.
2 here comes for the time of ascent and descent.
So,
So, the initial upward speed of the ball is 15.68 m/s.
d. all of these
look at the picture to know from where we get this law with application on it
Answer:
8.9
Explanation:
We can start by calculating the initial elastic potential energy of the spring. This is given by:
(1)
where
k = 35.0 N/m is the initial spring constant
x = 0.375 m is the compression of the spring
Solving the equation,
Later, the professor told the student that he needs an elastic potential energy of
U' = 22.0 J
to achieve his goal. Assuming that the compression of the spring will remain the same, this means that we can calculate the new spring constant that is needed to achieve this energy, by solving eq.(1) for k:
Therefore, Tom needs to increase the spring constant by a factor: