Answer:
Moment of inertia of the system is 289.088 kg.m^2
Explanation:
Given:
Mass of the platform which is a uniform disk = 129 kg
Radius of the disk rotating about vertical axis = 1.61 m
Mass of the person standing on platform = 65.7 kg
Distance from the center of platform = 1.07 m
Mass of the dog on the platform = 27.3 kg
Distance from center of platform = 1.31 m
We have to calculate the moment of inertia.
Formula:
MOI of disk = 
Moment of inertia of the person and the dog will be mr^2.
Where m and r are different for both the bodies.
So,
Moment of inertia 
of the system with respect to the axis yy.
⇒ 
⇒ 
⇒ 
⇒ 
The moment of inertia of the system is 289.088 kg.m^2
Answer:
<em>The balloon is 66.62 m high</em>
Explanation:
<u>Combined Motion
</u>
The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is
The values are
We must find the values of t such that the height of the camera is 0 (when it hits the ground)
Multiplying by 2
Clearing the coefficient of 
Plugging in the given values, we reach to a second-degree equation
The equation has two roots, but we only keep the positive root
Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is
Inertia. It also is the tendency of an object in motion to stay in motion in one specific direction.
The centripetal acceleration of an object is given by the relation,
where Ac = centripetal acceleration = 
R = radius of rotation = 15 m
V = speed of astronaut
Hence, 
solving this we get, V = 38.34 m/s
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows:
