Answer:
125.66 R/s
Explanation:
First 1200 r / min = 20 r/sec
20 r/s * 2pi Radians / r = 40 pi Radians / sec = 125.66 R/s
Answer:
a) 1.3 rad/s
b) 0.722 s
Explanation:
Given
Initial velocity, ω = 0 rad/s
Angular acceleration of the wheel, α = 1.8 rad/s²
using equations of angular motion, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
where
θ2 - θ1 = 53.2 rad
t2 - t1 = 7s
substituting these in the equation, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²
53.2 = 7.ω(0) + 1/2 * 1.8 * 49
53.2 = 7.ω(0) + 44.1
7.ω(0) = 53.2 - 44.1
ω(0) = 9.1 / 7
ω(0) = 1.3 rad/s
Using another of the equations of angular motion, we have
ω(0) = ω(i) + α*t1
1.3 = 0 + 1.8 * t1
1.3 = 1.8 * t1
t1 = 1.3/1.8
t1 = 0.722 s
The frequencies are missing in the question. The three successive resonance frequencies within the volume are 
.
Solution :
Let the volume be : v
The frequency for one end open and one end closed is given by :
So, 
Therefore,
, 
, 
So, 
and 
Therefore, the ratio of 
which is not a whole number.
Now the frequency for open volume and closed at both end
So, 
, 
, 
From above formulae we can see that ratio of is not a whole number that is a identification for the frequency of the volume at one end open and one end closed.
Also ratio of the consecutive frequency of the volume at open from both side and closed from both side is always a whole number.
Explanation:
The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end. The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.
Sum the torques about the wedge (it may help to draw a diagram first). Take counterclockwise to be positive.
∑τ = Iα
W (0.25 m) − (100 N) (0.50 m) = 0
W = 200 N
Sum the forces in the y direction.
∑F = ma
F − 100 N − 200 N = 0
F = 300 N
