Answer:
1) 1.71 rad/s
2) -6.22 rad/s²
Explanation:
Choose point C to be the origin.
Using geometry, we can show that the coordinates of point A are:
(a cos 30°, a sin 30° − b)
Therefore, the coordinates of point D at time t are:
(a cos 30° − b sin(ωt), a sin 30° − b + b cos(ωt))
The angle formed by CB with the x-axis is therefore:
tan θ = (a sin 30° − b + b cos(ωt)) / (a cos 30° − b sin(ωt))
1) Taking the derivative with respect to time, we can find the angular velocity:
sec² θ dθ/dt = [(a cos 30° − b sin(ωt)) (-bω sin(ωt)) − (a sin 30° − b + b cos(ωt)) (-bω cos(ωt))] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω [(a cos 30° − b sin(ωt)) sin(ωt) − (a sin 30° − b + b cos(ωt)) cos(ωt)] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω [(a cos 30° sin(ωt) − b sin²(ωt)) − (a sin 30° cos(ωt) − b + b cos²(ωt))] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − b sin²(ωt) − a sin 30° cos(ωt) + b − b cos²(ωt)) / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − a sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -abω (cos 30° sin(ωt) − sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²
We know at the moment shown, a = 350 mm, b = 200 mm, θ = 30°, ω = 6 rad/s, and t = 0 s.
sec² 30° dθ/dt = -(350) (200) (6) (cos 30° sin(0) − sin 30° cos(0)) / (350 cos 30° − 200 sin(0))²
sec² 30° dθ/dt = -(350) (200) (6) (-sin 30°) / (350 cos 30°)²
dθ/dt = (200) (6) (1/2) / 350
dθ/dt = 600 / 350
dθ/dt = 1.71 rad/s
2) Taking the second derivative of θ with respect to time, we can find the angular acceleration.
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30° − b sin(ωt))² (ω cos 30° cos(ωt) + ω sin 30° sin(ωt)) − (cos 30° sin(ωt) − sin 30° cos(ωt)) (2 (a cos 30° − b sin(ωt)) (-bω cos(ωt)))] / (a cos 30° − b sin(ωt))⁴
At t = 0:
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30°)² (ω cos 30°) − (0 − sin 30°) (2 (a cos 30°) (-bω))] / (a cos 30°)⁴
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω (a²ω cos³ 30° − 2abω sin 30° cos 30°) / (a⁴ cos⁴ 30°)
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -bω (aω cos² 30° − 2bω sin 30°) / (a² cos³ 30°)
d²θ/dt² + 2 tan θ dθ/dt = -bω² (a cos² 30° − b) / (a² cos 30°)
Plugging in values:
d²θ/dt² + 2 tan 30° dθ/dt = -(200) (6)² (350 cos² 30° − 200) / (350² cos 30°)
d²θ/dt² + 2 tan 30° dθ/dt = -7200 (262.5 − 200) / (350² cos 30°)
d²θ/dt² + 2 tan 30° (1.71) = -4.24
d²θ/dt² = -6.22 rad/s²
Answer:
The final pressure is 3.16 torr
Solution:
As per the question:
The reduced pressure after drop in it, P' = 3 torr =
At the end of pumping, temperature of air,
After the rise in the air temperature,
Now, we know the ideal gas eqn:
PV = mRT
So
(1)
where
P = Pressure
V = Volume
R = Rydberg's constant
T = Temperature
Using eqn (1):
Now, at constant volume the final pressure, P' is given by:
Answer:
a) 2.18 m/s^2
b) 9.83 m/s
Explanation:
The flywheel has a moment of inertia
J = m * k^2
Where
J: moment of inertia
k: radius of gyration
In this case:
J = 144 * 0.45^2 = 29.2 kg*m^2
The block is attached through a wire that is wrapped around the wheel. The weight of the block causes a torque.
T = p * r
r is the radius of the wheel.
T = m1 * g * r
T = 18 * 9.81 * 0.6 = 106 N*m
The torque will cause an acceleration on the flywheel:
T = J * γ
γ = T/J
γ = 106/29.2 = 3.63 rad/s^2
SInce the block is attached to the wheel the acceleration of the block is the same as the tangential acceleration at the eddge of the wheel:
at = γ * r
at = 3.63 * 0.6 = 2.81 m/s^2
Now that we know the acceleration of the block we can forget about the flywheel.
The equation for uniformly accelerated movement is:
X(t) = X0 + V0*t + 1/2*a*t^2
We can set a frame of reference that has X0 = 0, V0 = 0 and the X axis points in the direction the block will move. Then:
X(t) = 1/2*a*t^2
Rearranging
t^2 = 2*X(t)/a
It will reach the 1.8 m in 3.6 s.
Now we use the equation for speed under constant acceleration:
V(t) = V0 + a*t
V(3.6) = 2.81 * 3.6 = 9.83 m/s