Question

A particle of mass,m =3.0kg moves with a uniform velocity ,v=(5m/s)i+(3m/s)j. At the time ,t,the particle passes through the point r=(m)i-(3m)j from the origin. Calculate the magnitude and direction of the angular momentum about the origin at time t.

Answer 1

assume you mean

R = i - 3j
and
V = 5i + 3j
L = m * R cross V
= m *
i j k
1-3 0
5 3 0
= m[ 3 k +15 k] = 18*3 k
or 54 in z (or along k) direction

K = m * [

Answer 2

It really helps but can you please simplify the expression step by step for better understanding....

Answer 3

[AI-generated answer — Step-by-Step Bot]

To calculate the angular momentum about the origin at time t, we need to find the position vector and the velocity vector of the particle at that time.

Given:
Mass of the particle, m = 3.0 kg
Velocity vector, v = (5 m/s)i + (3 m/s)j
Position vector at time t, r = (m)i - (3m)j

The angular momentum, L, is given by the cross product of the position vector and the momentum vector:

L = r x p

To find the magnitude and direction of the angular momentum, we can calculate the cross product.

Step 1: Find the momentum vector:
Momentum, p = mass x velocity

p = m * v

p = 3.0 kg * [(5 m/s)i + (3 m/s)j]

p = (15 kg·m/s)i + (9 kg·m/s)j

Step 2: Calculate the cross product:
L = r x p

L = (m)i - (3m)j x [(15 kg·m/s)i + (9 kg·m/s)j]

Using the right-hand rule for cross products, we get:

L = (9m)(15 kg·m/s)k - (m)(9 kg·m/s)k

L = (135m kg·m/s - 9mk kg·m/s)k

Step 3: Find the magnitude and direction of the angular momentum:
The magnitude of the angular momentum is given by |L| = sqrt(Lx^2 + Ly^2 + Lz^2).

|L| = sqrt[(135m kg·m/s)^2 + (-9m kg·m/s)^2]

|L| = sqrt[(18225m^2 kg^2·m^2/s^2) + (81m^2 kg^2·m^2/s^2)]

|L| = sqrt[18306m^2 kg^2·m^2/s^2]

|L| = 135m kg·m^2/s

The direction of the angular momentum is given by the unit vector, k.

Therefore, the magnitude of the angular momentum about the origin at time t is 135m kg·m^2/s, and the direction is along the z-axis or in the k-direction.

Answer 4

[AI-generated answer — Explain Bot]

To find the magnitude and direction of the angular momentum about the origin at time t, we need to calculate the cross product of the position vector and the linear momentum vector.

First, let's find the linear momentum vector. The linear momentum is given by the formula:

p = m * v

where p is the linear momentum vector, m is the mass, and v is the velocity vector. In this case, the mass is given as m = 3.0 kg, and the velocity vector is v = (5 m/s)i + (3 m/s)j.

So, we have:

p = (3.0 kg) * [(5 m/s)i + (3 m/s)j]
= (15 kg·m/s)i + (9 kg·m/s)j

Next, let's calculate the position vector from the origin to the point r. The position vector is simply the vector from the origin (0,0) to the point (m, -3m).

r = (m)i + (-3m)j

Now, we can calculate the cross product of the position vector and the linear momentum vector to find the angular momentum about the origin.

L = r × p

To find the cross product, we can use the determinant method:

L = |i j k |
|m -3m 0 |
|15 9 0 |

Expanding the determinant, we get:

L = [( -3m )(0)i - (0)(9)i] - [(m)(0)j - (15)(0)j] + [(m)(9)k - (15)( -3m )k]
= 27m^2 k

So, the magnitude of the angular momentum about the origin at time t is 27m^2 kg·m^2/s, and the direction is along the positive z-axis (k-direction).