A 0.26 kg particle moves in an xy plane according to x(t) = - 13 + 1 t - 3 t3 and y(t) = 22 + 6 t - 11 t2, with x and y in meters and t in seconds. At t = 1.2 s, what are (a) the magnitude and (b) the angle (within (-180°, 180°] interval relative to the positive direction of the x axis) of the net force on the particle, and (c) what is the angle of the particle's direction of travel?
To find the net force on the particle at t = 1.2 s, we need to differentiate the position functions x(t) and y(t) to get the velocity functions vx(t) and vy(t). Then we can differentiate again to obtain the acceleration functions ax(t) and ay(t). Finally, we can use the acceleration components ax(t) and ay(t) to calculate the net force components and then find their magnitude and angles.
Let's start by finding the velocity components vx(t) and vy(t). The velocity of a particle is the derivative of its position with respect to time.
Given:
x(t) = -13 + t - 3t^3
y(t) = 22 + 6t - 11t^2
Differentiating the position functions with respect to time, we get:
vx(t) = d/dt [-13 + t - 3t^3] = 1 - 9t^2
vy(t) = d/dt [22 + 6t - 11t^2] = 6 - 22t
Next, let's find the acceleration components ax(t) and ay(t). The acceleration of a particle is the derivative of its velocity with respect to time.
Differentiating the velocity functions with respect to time, we get:
ax(t) = d/dt [1 - 9t^2] = -18t
ay(t) = d/dt [6 - 22t] = -22
Now we can calculate the net force components using Newton's second law, F = m * a, where m is the mass of the particle.
Given the mass of the particle is 0.26 kg, and at t = 1.2 s:
m = 0.26 kg
t = 1.2 s
Substituting these values into the acceleration components, we get:
ax(1.2) = -18 * 1.2 = -21.6 m/s^2
ay(1.2) = -22 m/s^2
Now we have the net force components as:
Fx = m * ax(1.2)
Fy = m * ay(1.2)
Calculating the net force components:
Fx = 0.26 kg * -21.6 m/s^2 = -5.616 N
Fy = 0.26 kg * -22 m/s^2 = -5.72 N
(a) To find the magnitude of the net force, we can use the Pythagorean theorem:
|F| = sqrt(Fx^2 + Fy^2)
Calculating the magnitude of the net force:
|F| = sqrt((-5.616 N)^2 + (-5.72 N)^2) = sqrt(62.986976 + 32.8384) = sqrt(95.825376) ≈ 9.79 N
Therefore, the magnitude of the net force on the particle is approximately 9.79 N.
(b) To find the angle of the net force, we can use the arctan function:
angle = atan(Fy / Fx)
Calculating the angle of the net force:
angle = atan((-5.72 N) / (-5.616 N)) = atan(1.02027) ≈ 46.33°
The angle of the net force on the particle is approximately 46.33° (measured counterclockwise from the positive direction of the x-axis).
(c) To find the angle of the particle's direction of travel, we can use the velocity components vx(t) and vy(t) at t = 1.2 s:
direction_angle = atan(vy(1.2) / vx(1.2))
Calculating the angle of the particle's direction of travel:
direction_angle = atan((-22 m/s) / (1 - 9(1.2)^2)) ≈ atan(-22 / (-9.48)) ≈ atan(2.320855) ≈ 67.19°
The angle of the particle's direction of travel is approximately 67.19° (measured counterclockwise from the positive direction of the x-axis).
To find the magnitude and angle of the net force on the particle, as well as the angle of the particle's direction of travel, we need to calculate the particle's acceleration at t = 1.2 s.
We can find the particle's acceleration by taking the second derivative of the position equations with respect to time:
a(t) = d^2x/dt^2 i + d^2y/dt^2 j
We first calculate the first derivatives of the position equations:
dx/dt = v_x(t) = d/dt (-13 + t - 3t^3) = 1 - 9t^2
dy/dt = v_y(t) = d/dt (22 + 6t - 11t^2) = 6 - 22t
Then we can find the second derivatives of the position equations:
d^2x/dt^2 = a_x(t) = d/dt (1 - 9t^2) = -18t
d^2y/dt^2 = a_y(t) = d/dt (6 - 22t) = -22
Calculating these derivatives, we have:
a_x(t) = -18t
a_y(t) = -22
Now we can find the values of acceleration at t = 1.2 s:
a_x(1.2) = -18(1.2) = -21.6 m/s^2
a_y(1.2) = -22 m/s^2
(a) To find the magnitude of the net force at t = 1.2 s, we can use the following equation:
F_net = sqrt((F_x)^2 + (F_y)^2)
where F_x and F_y are the x and y components of the net force.
F_x = m * a_x
F_y = m * a_y
Plugging in the values:
F_x = (0.26 kg) * (-21.6 m/s^2) = -5.616 N
F_y = (0.26 kg) * (-22 m/s^2) = -5.72 N
F_net = sqrt((-5.616 N)^2 + (-5.72 N)^2) = sqrt(31.617056 N^2 + 32.8084 N^2) ≈ 45.1 N
Therefore, the magnitude of the net force on the particle at t = 1.2 s is approximately 45.1 N.
(b) To find the angle of the net force relative to the positive direction of the x-axis, we can use the following equation:
θ = arctan(F_y / F_x)
θ = arctan((-5.72 N) / (-5.616 N)) = arctan(1.019) ≈ 46.7°
Since the angle needs to be within the (-180°, 180°] interval, this angle corresponds to 46.7° in the counter-clockwise direction from the positive x-axis.
(c) The angle of the particle's direction of travel can be found using the velocity components:
v_x(1.2) = 1 - 9(1.2)^2 = -11.68 m/s
v_y(1.2) = 6 - 22(1.2) = -18.4 m/s
θ_direction = arctan(v_y(1.2) / v_x(1.2)) = arctan((-18.4 m/s) / (-11.68 m/s)) = arctan(1.575) ≈ 56.5°
Therefore, the angle of the particle's direction of travel at t = 1.2 s is approximately 56.5°.