Complete Question
The complete Question is shown on the first uploaded image
Answer:
a
The torque acting on the particle is [tex]\tau = 48t \r k[/tex]
b
The magnitude of the angular momentum increases relative to the origin
Explanation:
From the equation we are told that
The position of the particle is [tex]\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j[/tex]
The mass of the particle is [tex]m = 3.0 kg[/tex]
The time is t
The torque acting on the particle is mathematically represented as
[tex]\tau = \frac{ d \r l }{dt}[/tex]
where [tex]\r l[/tex] is change in angular momentum which is mathematically represented as
[tex]\r l = m (\r r \ \ X \ \ \r v)[/tex]
Where X mean cross- product
[tex]\r v[/tex] is the velocity which is mathematically represented as
[tex]\r v = \frac{d \r r }{dt}[/tex]
Substituting for [tex]\r r[/tex]
[tex]\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j][/tex]
[tex]\r v = 8t \r i - (2 + 12 t) \r j[/tex]
Now the cross product of [tex]\r r \ and \ \r v[/tex] is mathematically evaluated as
[tex]\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right][/tex]
[tex]= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k[/tex]
[tex]\r r \ \ X \ \ \r v = 8t^2 \r k[/tex]
So the angular momentum becomes
[tex]\r l = m (8t^2 \r k)[/tex]
Substituting for m
[tex]\r l = 3 * (8t^2 \r k)[/tex]
[tex]\r l =24t^2 \r k[/tex]
Substituting into equation for torque
[tex]\tau = \frac{d}{dt} [24t^2 \r k][/tex]
[tex]\tau = 48t \r k[/tex]
The magnitude of the angular momentum can be evaluated mathematically as
[tex]|\r l| = \sqrt{(24 t^2) ^2}[/tex]
[tex]|\r l| = 24 t^2[/tex]
From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also
