Physics, 29.02.2020 08:40 Tink2334

At time t= 0, a 3.0 kg particle with velocity v= (5.0 m/s)i - (6.0 m/s)j is at x = 3.0 m, y= 8.0m. it is pulled by a 7.0 N force in the negative x direction. About the origin, what are the particle's angular momentum, the torque acting on the particle and the rate at which the angular momentum is changing?

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Answer from: ysabel0420

a) The angular momentum of the particle is $\vec L = -174\,\hat{k}\,\left[\frac{kg\cdot m}{s} \right]$ .

b) The torque acting on the particle is $\vec {\tau} = -56\,\hat{k}\,[N \cdot m]$ .

c) The rate at which the angular momentum is changing is $\vec {\tau} = -56\,\hat{k}\,[N \cdot m]$ .

a) The angular momentum of the particle ( $\vec L$ ), in kilograms-meter per second, is given by the following vectorial formula:

$\vec L = \vec r \,\times\, m\cdot \vec v$ (1)

Where:

$\vec r$ - Distance with respect to the origin, in meters.

- Mass, in kilograms. $\vec v$ - Velocity, in meters per second.

If we know that $\vec r = (3\,m, 8\,m, 0\,m)$ , $m = 3\,kg$ and $\vec v = \left(5\,\frac{m}{s}, -6\,\frac{m}{s}, 0\,\frac{m}{s} \right)$ , then the angular momentum of the particle is determined by determinants:

$\vec L = (3\,kg)\cdot \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\3\,m&8\,m&0\,m\\5\,\frac{m}{s} &-6\,\frac{m}{s} &0\,\frac{m}{s} \end{array}\right|$

$\vec L = (3\,kg)\cdot \left[(3\,m)\cdot \left(-6\,\frac{m}{s} \right)-(8\,m)\cdot \left(5\,\frac{m}{s} \right)\right]\cdot \hat{k}$

$\vec L = -174\,\hat{k}\,\left[\frac{kg\cdot m}{s} \right]$

The angular momentum of the particle is $\vec L = -174\,\hat{k}\,\left[\frac{kg\cdot m}{s} \right]$ .

b) The torque ( $\vec \tau$ ), in newton-meters, acting on the particle is defined by the following expression:

$\vec {\tau} = \vec {r}\,\times\,\vec {F}$ (2)

Where $\vec F$ is the force, in newtons, acting on the particle.

If we know that $\vec r = (3\,m, 8\,m, 0\,m)$ and $\vec F = (-7\,N, 0\,N, 0\,N)$ , then the torque experimented by the particle is determined by determinants:

$\vec {\tau} = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\3&8&0\\-7&0&0\end{array}\right| \,[N\cdot m]$

$\vec {\tau} = -56\,\hat{k}\,[N \cdot m]$

The torque acting on the particle is $\vec {\tau} = -56\,\hat{k}\,[N \cdot m]$ .

c) The rate at which the angular momentum changes in time is found deriving (1):

$\vec {\dot L} = \vec {v} \,\times \,m\cdot \vec {v} + \vec {r}\,\times\,m\cdot \vec {a}$

$\vec{\dot {L}} = \vec O + \vec r \,\times \,\vec F$

$\vec {\dot L} = \vec r \,\times \,\vec F$

$\vec {\dot L} = \vec{\tau}$ (3)

In other words, the rate of a constant mass system at which the angular momentum changes is equal to the torque experimented by the particle. Hence, the rate at which the angular momentum changes is $\vec {\tau} = -56\,\hat{k}\,[N \cdot m]$ .

The rate at which the angular momentum is changing is $\vec {\tau} = -56\,\hat{k}\,[N \cdot m]$ .

We kindly invite to check this question on torque: link

Answer from: kiekie1986

Explanation:

The radius, velocity, and force vectors are:

r = (3.0 m) i + (8.0 m) j

v = (5.0 m/s) i − (6.0 m/s) j

F = (-7.0 N) i

Angular momentum is the cross product of the radius vector and the linear momentum vector:

L = r × p

L = r × (mv)

L = <(3.0 m) i + (8.0 m) j> × <(15 kg m/s) i + (-18 kg m/s) j>

L = (-174 kg m²/s) k

Torque is the cross product of the radius vector and the force vector:

τ = r × F

τ = <(3.0 m) i + (8.0 m) j> × <(-7.0 N) i + (0 N) j>

τ = (56 Nm) k

Rate of change of angular momentum is equal to the torque.

L' = τ

L' = (56 Nm) k

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+10 that is always the common denominator

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At time t= 0, a 3.0 kg particle with velocity v= (5.0 m/s)i - (6.0 m/s)j is at x = 3.0 m, y= 8.0m. i...