Question

three bodies of masses 2kg,3kg and 5kg have position vector i+I,i-j and 2i+I respectively. if it has given velocity 3i+4j,2i+2j and -3i-4j respectively then angle between position of centre of mass and velocity of centre of mass will be​

Answer 1

Given that,

There are three bodies of mass 2kg, 3kg and 5kg having position vector i + j , i - j and 2i + j with velocity 3i + 4j , 2i + 2j and -3i - 4j respectively.

We have to find the angle between position vector of centre of mass and velocity vector of centre of mass.

Let's find Position vector of COM first. To find position vector of COM of bodies we just need to have the mass of those bodies as well as the position vector. This can be calculated using the given formula,

⇒ R = Σmₙrₙ / M

Where,

• Σmₙrₙ denotes the vector sum of product of mass m and the position vector r.
• M denotes the total sum of mass of the bodies.

So,

⇒ R = { 2(i + j) + 3(i - j) + 5(2i + j) } / (2 + 3 + 5)

⇒ R = ( 2i + 2j + 3i - 3j + 10i + 5j ) / (10)

[ ∵ n(Xi + Yj) = nXi + nYj , where, n = constant ]

⇒ R = { (2 + 3 + 10)i + (2 + (-3) + 5)j } / 10

⇒ R = ( 15i + 4j ) / 10

⇒ R = 15/10 i + 4/10 j

⇒ R = 3/2 i + 2/5 j

Similarly, Let's find the velocity vector of COM, which is given by the formula,

⇒ V = Σmₙvₙ / M

Where,

• Σmₙvₙ denotes the vector sum of product of mass and velocity vector of each body.
• M denotes the total sum of mass of the bodies.

Substituting the values, we get

⇒ V = { 2(3i + 4j) + 3(2i + 2j) + 5(- 3i - 4j) } / (2 + 3 + 5)

⇒ V = ( 6i + 8j + 6i + 6j - 15i - 20j ) / (10)

⇒ V = { (6 + 6 - 15)i + (8 + 6 - 20)j } / 10

⇒ V = ( -3i - 6j ) / 10

⇒ V = -3/10 i - 6/10 j

Now, We have got the required vectors, Let's find the angle using the given formula,

⇒ = cos⁻¹ [ R.V / |R| |V| ]

Where,

• R.V denotes the scalar product of Position vector and Velocity vector of COM.
• |R| |V| denotes the product of magnitudes of position vector and velocity vector of COM respectively.

Let's find the scalar product first,

We know, Scalar product of two vectors,

• A = Aₓi + Aᵧj + A₂k
• B = Bₓi + Bᵧj + B₂k

is given by,

⇒ A . B = AₓBₓ + AᵧBᵧ + A₂B₂

So,

⇒ R . V = (3/2 × -3/10) + (2/5 × -6/10)

⇒ R . V = -9/20 - 12/50

⇒ R . V = (-9×5 - 12×2) / 100

⇒ R . V = (-45 - 24) / 100

⇒ R . V = -69/100

Now, Let's find the magnitude of Position and Velocity vector of COM:

Position Vector:

⇒ |R| = √{ (3/2)² + (2/5)² }

⇒ |R| = √{ 9/4 + 4/25 }

⇒ |R| = √{ (25×9 + 4×4)/100 }

⇒ |R| = √( 241/100 )

⇒ |R| = √(241) / 10

Velocity Vector:

⇒ |V| = √{ (-3/10)² + (-6/10)² }

⇒ |V| = √{ 9/100 + 36/100 }

⇒ |V| = √{45/100}

⇒ |V| = 3√5 / 10

Now, Substitute the values:

⇒ = cos⁻¹ (-69/100) / (√241/10 × 3√5/10)

⇒ = cos⁻¹ (-69/√241×3√5)

⇒ = cos⁻¹ (-23/√1205)

⇒ = cos⁻¹ (-0.66257)

⇒ = 131.49° or ≈ 131.5°

Hence, The angle b/w the position vector and velocity vector of COM is ≈ 131.5°

Answer 2

131.5°

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