Answer:
-0.5 m/s2 as it comes to rest.
Explanation:
Answer:
ωf = 13 rad/s
Explanation:
- The angular acceleration, by definition, is just the rate of change of the angular velocity with respect to time, as follows:
- α = Δω/Δt = (ωf-ω₀) / (tfi-t₀)
- Choosing t₀ = 0, and rearranging terms, we have
where ω₀ = 5 rad/s, t = 4 s, α = 2 rad/s2
- Replacing these values in (1) and solving for ωf, we get:
- The wheel's angular velocity after 4s is 13 rad/s.
Answer:
4.5sec
Explanation:
From the question above, the following are the parameters that are given
u= 30m/s
v= 50m/s
s= 180m
First of all we have to find the acceleration by using the third equation of motion
V^2= U^2 + 2as
50^2= 30^2 + 2(a)(180)
2500= 900 + 360a
Collect the like terms
2500-900= 360a
1600=360a
Divide both sides by the coefficient of a which is 360
1600/360=360a/360
a= 4.44m/s
The next step is to find the time. To do this we will have to use the first equation of motion
v= u + at
50= 30 + 4.44t
Collect the like terms
50-30= 4.44t
20= 4.44t
Divide both sides by the coefficient of t which is 4.44
20/4.44= 4.44t/4.44
t= 4.5sec
Hence 4.5secs elapses while the auto moves at a distance of 180m
I found some good web pages with highly detailed answers to predicting the range of a trebuchet. A very simple model we have used in my Intro to Eng class just uses the mass of the projectile (m2), the mass of the counter weight (m1), and the height the counter weight falls (h):
Range (max) = 2 * (m1/m2) * h
Now the efficiency of the trebuchet will cause this model to be off by quite a bit. But once you have a working trebuchet, we find this model works well when we vary m1, m2, or h. We assume we have a take off angle of 45 degrees above the horizon.
This solution is based on the classic max range ballistics problem - 45 degree take off angle. It also assumes converting all the potential energy of the counter weight to kinetic energy of the projectile. That is why the efficiency issue comes up as a lot of energy is lost due to friction in the moving trebuchet. If the projectile spins a lot then it will travel a shorter distance as the potential energy is split into kinetic and rotational energy. Projectile shape and wind will also vary the results.
Good luck.
