A passenger on a train notices that a 2.00 kg mass is displaced 10.0° from the vertical as the train
accelerates. (a) What is the acceleration of the train? (b) What would the acceleration be if the
mass was 5.00 kg?
c. What would angle have to be to indicate a horizontal acceleration equal to the
acceleration of gravity?
d. Could the acceleration ever be great enough to have the angle be 90.0° -- that is,
have the cord suspending the mass be horizontal? Explain.
a. tan10=a/g
b. same
c. 45 deg
d. what is the tangent of 90 deg?
I am confused with c where did the 45 degrees come from
and also how do you know to use the tan of 10
To solve these problems, we can use the concepts of Newton's second law of motion and centripetal acceleration. Let's break down each question one by one:
(a) To calculate the acceleration of the train, we need to use the formula for centripetal acceleration:
a = (v^2) / r
In this case, since the mass is displaced from the vertical, we should consider the vertical component of acceleration. The vertical component of acceleration is given by:
a_vertical = a * sin(theta)
where theta is the angle of displacement (10.0° in this case). Since we are given the mass (2.00 kg), we can calculate the vertical component of acceleration:
a_vertical = (g * m) / (sqrt(1 + (g * r / v^2)^2)) * sin(theta),
where g is the acceleration due to gravity (9.8 m/s^2) and r is the radius of the circular path. In this case, since we don't have the radius, we can assume that the radius is large enough that r/v^2 ≈ 0, making the equation simpler:
a_vertical = g * sin(theta).
Substituting the values, we get:
a_vertical = 9.8 m/s^2 * sin(10.0°) ≈ 1.70 m/s^2.
So the acceleration of the train is approximately 1.70 m/s^2.
(b) To find the acceleration if the mass is 5.00 kg, we can use the same formula as before:
a_vertical = g * sin(theta).
Substituting the values, we get:
a_vertical = 9.8 m/s^2 * sin(10.0°) ≈ 1.70 m/s^2.
So the acceleration of the train would still be approximately 1.70 m/s^2, regardless of the mass of the object.
(c) To determine the angle that would indicate a horizontal acceleration equal to the acceleration of gravity, we can set the vertical component of acceleration equal to zero:
a_vertical = g * sin(theta) = 0.
Solving for theta, we get:
sin(theta) = 0.
The only angle that satisfies this condition is theta = 0°, which means the cord suspending the mass would be vertical.
(d) No, the acceleration cannot be great enough to have the angle be 90.0° and have the cord suspending the mass be horizontal. The maximum angle that can be achieved is 90.0° - theta. Having the angle as 90.0° would mean that the entire gravitational force is canceled by the tension in the cord, resulting in zero tension and no net force acting on the mass. Therefore, the cord would not be able to suspend the mass.
In summary:
(a) The acceleration of the train is approximately 1.70 m/s^2.
(b) The acceleration would still be approximately 1.70 m/s^2 regardless of the mass.
(c) The angle needed to indicate a horizontal acceleration equal to gravity is 0°.
(d) No, the acceleration cannot be great enough for the cord to be horizontal.