The kinetic energy of the pendulum bob in figure 15-1 increases the most between locations
2 answers:
Answer:
A and C
Explanation:
Kinetic energy of a body is given by the expression, , where v is the velocity and m is the mass of body.
In case of a simple pendulum the maximum velocity is at it's stationary position.
So maximum velocity i at position C, which means maximum kinetic energy is at position A.
The velocity of pendulum reduces as the angle between vertical stationary and current position increases.
So, velocity at E and A are the minimum, so at A and E the kinetic energy value is minimum.
Now examining the options given, we will understand the kinetic energy increase is maximum in case of A and C.
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Answer:
a) T_A = , b) T_B = g [m₂ (
) + m₁ (
) ]
c) x = , d) m₂ = m₁ (
)
Explanation:
After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act
a) The tension of string A is requested
The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive
∑ τ = 0
T_A d - W₂ x -W₁ L/2 = 0
T_A =
b) the tension in string B
we write the expression of the translational equilibrium
∑ F = 0
T_A - W₂ - W₁ - T_B = 0
T_B = T_A -W₂ - W₁
T_ B = - g m₂ - g m₁
T_B = g [m₂ ( ) + m₁ (
) ]
c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system
T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0
at the point that begins to rotate T_B = 0
g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0
m₂ (d-x) = m₁ (0.5 L- d)
m₂ x = m₂ d - m₁ (0.5 L- d)
x =
d) The mass of the block for which it is always in equilibrium
this is the mass for which x = 0
0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}
m₂ = m₁
m₂ = m₁ ( )
