At time t, r^=4t^2 i^-(2t+6t^2)j^ gives the position of a particle of mass m relative to the origin of an xy coordinate system Find an expression for the (a) x, (b) y, and (c) z components of the torque acting on the particle relative to the origin.
since motion is in x,y plane, torque is about z axis
what is linear acceleration?
dr/dt = 8 t i -( 2+12 t)j
d^2r/dt^2 = 8 i -12 j constant linear acceleration so constant linear force
F = m (8 i - 12 j) = 4m (2 i - 3j)
now what is F cross r , the torque
i j k
rx ry 0
2 -3 0
all times 4 m
4m (-3rx -2 ry)
in k which is z direction
4 m [ -3(4t^2) +2(2t+6t^2)]
4 m [ 12 t^2 - 12 t^2 + 4 t]
16 m t
sorry, sign reversed
moment = R cross F not F cross R
it has been a while
check all my arithmetic
To find the expression for the x and y components of the torque acting on the particle, we first need to calculate the position vector, velocity vector, and the acceleration vector.
Given that the position vector r is given as:
r = 4t^2i - (2t + 6t^2)j
We can see that the position vector r has both x and y components. The z-component is not given in the equation, so we can assume it to be zero since it is relative to the origin in an xy coordinate system.
(a) x-component:
To find the x-component, we can simply extract the coefficient of the i unit vector. In this case, it is 4t^2. Therefore, the x-component of the position vector is given by:
x = 4t^2
(b) y-component:
Similarly, to find the y-component, we extract the coefficient of the j unit vector. In this case, it is -(2t + 6t^2). Therefore, the y-component of the position vector is given by:
y = -(2t + 6t^2)
(c) z-component:
The z-component is assumed to be zero as mentioned earlier.
To calculate the velocity vector, we need to differentiate the position vector with respect to time (t).
v = dr/dt
Differentiating each component of the position vector:
dx/dt = d(4t^2)/dt = 8t
dy/dt = d(-(2t + 6t^2))/dt = -2 - 12t
So, the velocity vector v is given by:
v = 8ti - (2 + 12t)j
To calculate the acceleration vector, we differentiate the velocity vector with respect to time (t).
a = dv/dt
Differentiating each component of the velocity vector:
d^2x/dt^2 = d(8t)/dt = 8
d^2y/dt^2 = d(-(2 + 12t))/dt = -12
So, the acceleration vector a is given by:
a = 8i - 12j
Now that we have the position vector, velocity vector, and acceleration vector, we can calculate the torque acting on the particle relative to the origin using the cross product of the position vector with the force vector.
Torque (τ) = r x m * a
As we are given the mass (m) is not mentioned, we cannot directly calculate the torque. Mass m is needed to calculate the torque.