Q35P Question: At time t, the vector ... [FREE SOLUTION]

91影视

Chapter 11: Q35P (page 322)

Question: At time t, the vector $\overset{鈫�}{r} = 4.0 t^{2} \hat{i} - (2.0 t + 6.0 t^{2}) \hat{j}$ gives the position of a3 .0 kgparticle relative to the origin of ancoordinate system (is in meters and tis in seconds). (a) Find an expression for the torque acting on the particle relative to the origin. (b) Is the magnitude of the particle鈥檚 angular momentum relative to the origin increasing, decreasing, or unchanging?

Short Answer

Expert verified

Answer

a.The torque acting on the particle relative to the origin is $\overset{鈫�}{蟿} = 48 t \hat{k}$

b.The magnitude of the particle鈥檚 angular momentum relative to origin is increasing

Step by step solution

Given

The mass of the particle is $m = 3.0 k g$

The position vector is $\overset{鈫�}{r} = 4.0 t^{2} \hat{i} - (2.0 t + 6.0 t^{2}) \hat{j}$

To understand the solution

Find the torque acting on the particle by using concept of Newton鈥檚 second law in angular form.

Formula:

$\frac{d \overset{鈫�}{l}}{d t} = {\overset{鈫�}{蟿}}_{n e t}$

Calculate the torque acting on particle relative origin

(a)

The expression of the velocity is,

$\overset{鈫�}{v} = \frac{d \overset{鈫�}{r}}{d t}$

Putting the position vector in above equation

role="math" localid="1661255607475" $\overset{鈫�}{v} = \frac{d [4.0 t^{2} \hat{i} - (2.0 t + 6.0 t^{2}) \hat{j}]}{d t} 鈬� \overset{鈫�}{v} = 8.0 t \hat{i} - (2.0 + 12 t) \hat{j}$

Let position vector be $\overset{鈫�}{r} = x \hat{i} + y \hat{j} + z \hat{k}$ and velocity vector be $\overset{鈫�}{v} = v_{x} \hat{i} + v_{y} \hat{j} + v_{z} \hat{k}$ .

The cross product of the position vector and velocity vector is

$\overset{鈫�}{r} 脳 \overset{鈫�}{v} = (y v_{z} - z v_{y}) \hat{i} + (z v_{x} - x v_{z}) \hat{j} + (x v_{y} - y v_{x}) \hat{k}$

In the given position and velocity vector, z = 0 m and $v_{z} = 0 m / s$ . Then

$\overset{鈫�}{r} 脳 \overset{鈫�}{v} = (x v_{y} - y v_{x}) \hat{k}$

The angular momentum of the object with position vector and velocity vector is

$\overset{鈫�}{l} = m (\overset{鈫�}{r} 脳 \overset{鈫�}{v}) 鈬� \overset{鈫�}{l} = m (x v_{y} - y v_{x}) \hat{k} 鈬� \overset{鈫�}{l} = 3.0 k g [(4.0 t^{2}) (- 2.0 - 12 t) + (2.0 t + 6.0 t^{2}) (8.0 t)] \hat{k} 鈬� \overset{鈫�}{l} = 3.0 k g (- 8.0 t^{2} - 48 t^{3} + 16 t^{2} + 48 t^{3})$

According to Newton鈥檚 second law in angular form, the sum of all torques acting on a particle is equal to the time rate of the change of the angular momentum of that particle.

$鈬� \overset{鈫�}{蟿} = \frac{d \overset{鈫�}{l}}{d t} 鈬� \overset{鈫�}{蟿} = (24 \hat{k}) \frac{d t^{2}}{d t} 鈬� \overset{鈫�}{蟿} = 48 t \hat{k}$