Answer:
1)m=89.01 g
2)V(max) = 97.3 cm/s
Explanation:
Given that
K= 15 N/m
The maximum amplitude ,A=7.5 cm = 0.075 m
Given that 31 oscillations in 15 seconds ,this means that frequency f
f=\dfrac{31}{15}
f=2.066 Hz
lets take mass of the ball = m kg
m=0.08901 kg
m=89.01 g
The maximum speed
V(max)= ω x A
ω = 2π f= 2 x π x 2.066 = 12.98 rad/s
V(max) = 12.98 x 0.075 =0.973 m/s
V(max) = 97.3 cm/s
Answer:
v = 10 [m/s]
Explanation:
The largest mass is that of 4 [kg], in this way the momentum can be calculated by means of the product of the mass by velocity.
where:
P = momentum [kg*m/s]
m = mass = 4 [kg]
v = velocity = 5 [m/s]
Now the momentum:
This same momentum is equal for the other mass, in this way we can find the velocity.
Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m