Answer:
a)0.0229 m
b)0.393 rad
c)1.57
d)707.6 N
e)0.298 m/s
Explanation:
Given:
- Mass of the machine, m=70 kg
- Stiffness of the system, k=30000 N/m
- Damping ratio=0.2
- Damping force, F=450 N
- Angular velocity
a)We know that the amplitude X at steady state is given by
Where
![X=\dfrac{\dfrac{450}{70}}{\sqrt{20.7^2-13^2)^2 +(2\times 0.2\times20.7\times13)^2}}\\[tex]X=0.0229\ \rm m](https://tex.z-dn.net/?f=X%3D%5Cdfrac%7B%5Cdfrac%7B450%7D%7B70%7D%7D%7B%5Csqrt%7B20.7%5E2-13%5E2%29%5E2%20%2B%282%5Ctimes%200.2%5Ctimes20.7%5Ctimes13%29%5E2%7D%7D%5C%5C%5Btex%5DX%3D0.0229%5C%20%5Crm%20m)
b) The phase shift of the motion is given by
c)Transmissibility ratio is given by
d)The magnitude of the force transmitted to the ground is
e)The maximum velocity is given by 
95-2.30+0.1+0.02cm =0.00089186
100m
Answer:
1.) Time t = 3.1 seconds
2.) Height h = 46 metres
Explanation:
given that the initial velocity U = 30 m/s
At the top of the trajectory, the final velocity V = 0
Using first equation of motion
V = U - gt
g is negative 9.81m/^2 as the object is going against the gravity.
Substitute all the parameters into the formula
0 = 30 - 9.81t
9.81t = 30
Make t the subject of formula
t = 30/9.81
t = 3.058 seconds
t = 3.1 seconds approximately
Therefore, it will take 3.1 seconds to reach to reach the top of its trajectory.
2.) The height it will go can be calculated by using second equation of motion
h = ut - 1/2gt^2
Substitutes U, g and t into the formula
h = 30(3.1) - 1/2 × 9.8 × 3.1^2
h = 93 - 47.089
h = 45.911 m
It will go 46 metres approximately high.
