At time t = 0, a particle of mass m has position vector (r^=4i-2j^ ) relative to the origin, in meters. For t ≥ 0, its velocity is given by v^=-6t^2i^+4j^.Find expressions of the z component of (a) the particle's angular momentum and (b) the torque acting on the particle at time t.
t is in s and v in m/s
someone pls help ;)
angular momentum L = R cross P = R cross m V
R(0) = 4 i - 2 j at t = 0
R(t) = R(0) + integral V dt from t = 0 to t = t
R(t) = 4 i - 2j -6 integral t^2 dt i^+4 integral dt j
R(t) = (4-2t^3)i + 4 t j
here P = m V = m [ -6 t^2 i + 4 j]
so L =R cross mV
L =m [ 4(4-2t^3)+24t^3] k = m [16-8t^3+24 t^3]
L =16 m[ 1-t^3]
and then
torque = d/dt L
= m[-48 t^2 ]
once again check my arithmetic, I stand behind method only
Are you at MIT? (That sort of problem)
Thank you for help, I’m not at MIT I’m studying engineering tho.
To find the z component of the particle's angular momentum (Lz) at time t, we can use the formula:
Lz = r x p
where "x" represents the cross product and "p" represents the linear momentum of the particle.
First, let's calculate the linear momentum (p) of the particle. The linear momentum is the product of the particle's mass (m) and its velocity (v). Given that the particle's velocity is v^ = -6t^2i^ + 4j^, we can find the linear momentum as follows:
p = m * v
= m * (-6t^2i^ + 4j^)
Next, we find the cross product between the position vector (r) and the linear momentum (p). The cross product of two vectors in 2D can be calculated as:
r x p = (rx * py - ry * px)k^
where "rx" and "ry" are the x and y components of the position vector, and "px" and "py" are the x and y components of the linear momentum.
The k^ is the unit vector in the z-direction to denote the z component of the vector.
Given r^ = 4i^ - 2j^ and p = -6t^2i^ + 4j^, we can substitute these values into the formula to find the z component of the angular momentum (Lz):
Lz = (4 * 4 - (-2) * (-6t^2)) k^
Simplifying the expression, we get:
Lz = (16 + 12t^2) k^
Therefore, the expression for the z component of the particle's angular momentum is (16 + 12t^2) k^.
Now, let's find the torque acting on the particle at time t. The torque (τ) is given by the cross product of the position vector (r) and the force vector (F):
τ = r x F
Since the force acting on the particle is equal to the rate of change of its linear momentum, we can express F as:
F = dp/dt
To find F, we differentiate the momentum (p) with respect to time (t):
dp/dt = m * dv/dt
= (-6m)(2t)i^
Now, using the cross product formula, we can find the z component of torque (τz):
τz = (4 * (-6m)(2t) - (-2) * 0) k^
Simplifying the expression, we get:
τz = -48m't k^
Therefore, the expression for the torque acting on the particle at time t is -48m't k^, where m' is the effective mass, which is m multiplied by any moment of inertia factor required by the problem.