A toy rocket is launched with an initial velocity of 10.0 m/s in the horizontal direction from the roof of a 38.0-m-tall buildin
1 answer:
You might be interested in
Answer:
a) The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: ,
.
b) Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: ,
.
c) The Ferris wheel takes 14.646 seconds to make a revolution.
Explanation:
a) An object that rotates at constant angular velocity reports a centripetal acceleration and no tangential acceleration. When passenger passes through the lowest point in her circular motion, centripetal acceleration goes up to the center.
In addition, centripetal acceleration is determined by the following expression:
(Eq. 1)
Where:
- Centripetal acceleration, measured in meters per square second.
- Linear speed, measured in meters per second.
- Radius of the Ferris wheel, measured in meters.
If we know that and
, the magnitude of radial acceleration is:
The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: ,
.
b) In the highest point the magnitude of radial acceleration is the same but direction is the opposed to that at lowest point. That is, centripetal acceleration goes down to the center.
Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: ,
.
c) At first we need to calculate the angular velocity of the Ferris wheel (), measured in radians per second, by using the following expression:
(Eq. 2)
If we know that and
, then the angular velocity of the Ferris wheel is:
Now we proceed to obtain the period of the Ferris wheel (), measured in seconds, which is the time needed by that wheel to make on revolution:
(Eq. 3)
Where is the angular velocity of the Ferris wheel, measured in radians per second.
If we get that , then:
The Ferris wheel takes 14.646 seconds to make a revolution.
The block on the action of two forces, the Force of Friction and the Tangential Weight. Using the Newton's Secound Law, we have:
Using the Velocity Hourly Equation, we get:
Uniting the equations:
Entering the unknowns:
Obs: Approximate results
If you notice any mistake in my english, please let me know, because i am not native.
Using the Velocity Hourly Equation, we get:
Uniting the equations:
Entering the unknowns:
Obs: Approximate results
If you notice any mistake in my english, please let me know, because i am not native.