Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Answer:
Average speed = 3.63 m/s
Explanation:
The average speed during any time interval is equal to the total distance travelled divided by the total time.
That is,
Average speed = distance/ time
Let d represent the distance between A and B.
Let t1 be the time for which she has the higher speed of 5.15 m/s. Therefore,
5.15 = d/t1.
Make d the subject of formula
d = 5.15t1
Let t2 represent the longer time for the return trip at 2.80 m/s . That is,
2.80 = d/t2.
Then the times are t1 = d/5.15 5 and
t2 = d/2.80.
The average speed vavg is given by the following equation.
avg speed = Total distance/Total time
Avg speed = d + d/t1 + t2
Where
Total distance = 2d
Total time = t1 + t2
Total time = d/5.15 + d/2.80
Total time = (2.8d + 5.15d)/14.42
Total time = 7.95d/14.42
Total time = 0.55d
Substitute total distance and time into the formula above.
Avg speed = 2d / 0.55d
Avg Speed = 3.63 m/s
Answer:
angular velocity(ω) is the rate change of angular displacement.
ω=θ/t and it SI unit is rad/s
Explanation:
this is very similar with the definition of linear velocity (rate of change of displacement). it specifies the angular speed of an object and the axis about which the object is rotating.
Answer:
ω = 12.023 rad/s
α = 222.61 rad/s²
Explanation:
We are given;
ω0 = 2.37 rad/s, t = 0 sec
ω =?, t = 0.22 sec
α =?
θ = 57°
From formulas,
Tangential acceleration; a_t = rα
Normal acceleration; a_n = rω²
tan θ = a_t/a_n
Thus; tan θ = rα/rω² = α/ω²
tan θ = α/ω²
α = ω²tan θ
Now, α = dω/dt
So; dω/dt = ω²tan θ
Rearranging, we have;
dω/ω² = dt × tan θ
Integrating both sides, we have;
(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ
This gives;
-1[(1/ω_o) - (1/ω)] = t(tan θ)
Thus;
ω = ω_o/(1 - (ω_o × t × tan θ))
While;
α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²
Thus, plugging in the relevant values;
ω = 2.37/(1 - (2.37 × 0.22 × tan 57))
ω = 12.023 rad/s
Also;
α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²
α = 8.64926751525/0.03885408979 = 222.61 rad/s²