Step 1: Identify the given parameters.
The position vector of the particle of mass m = 0.1 kg m = 0.1\,\text{kg} m = 0.1 kg
r ⃗ ( t ) = α t 3 i ^ + β t 2 j ^ \vec{r}(t) = \alpha t^{3}\hat{i} + \beta t^{2}\hat{j} r ( t ) = α t 3 i ^ + β t 2 j ^ where the constants are:
α = 10 3 ms − 3 , β = 5 ms − 2 \alpha = \frac{10}{3}\,\text{ms}^{-3},\quad \beta = 5\,\text{ms}^{-2} α = 3 10 ms − 3 , β = 5 ms − 2 Step 2: Find the velocity vector and evaluate at t = 1 s t = 1\,\text{s} t = 1 s
Differentiate the position vector with respect to time t t t
v ⃗ ( t ) = d r ⃗ d t = 3 α t 2 i ^ + 2 β t j ^ \vec{v}(t) = \frac{d\vec{r}}{dt} = 3\alpha t^{2}\hat{i} + 2\beta t\hat{j} v ( t ) = d t d r = 3 α t 2 i ^ + 2 β t j ^ Substitute t = 1 s t = 1\,\text{s} t = 1 s α = 10 3 ms − 3 \alpha = \frac{10}{3}\,\text{ms}^{-3} α = 3 10 ms − 3 β = 5 ms − 2 \beta = 5\,\text{ms}^{-2} β = 5 ms − 2
v ⃗ ( 1 ) = 3 ( 10 3 ) ( 1 ) 2 i ^ + 2 ( 5 ) ( 1 ) j ^ \vec{v}(1) = 3\left(\frac{10}{3}\right)(1)^{2}\hat{i} + 2(5)(1)\hat{j} v ( 1 ) = 3 ( 3 10 ) ( 1 ) 2 i ^ + 2 ( 5 ) ( 1 ) j ^ v ⃗ ( 1 ) = 10 i ^ + 10 j ^ ms − 1 \vec{v}(1) = 10\hat{i} + 10\hat{j}\ \text{ms}^{-1} v ( 1 ) = 10 i ^ + 10 j ^ ms − 1 Thus, Option (1) is TRUE .
Step 3: Find the acceleration and force vectors at t = 1 s t = 1\,\text{s} t = 1 s
Differentiate the velocity vector with respect to time t t t
a ⃗ ( t ) = d v ⃗ d t = 6 α t i ^ + 2 β j ^ \vec{a}(t) = \frac{d\vec{v}}{dt} = 6\alpha t\hat{i} + 2\beta\hat{j} a ( t ) = d t d v = 6 α t i ^ + 2 β j ^ Evaluate the acceleration at t = 1 s t = 1\,\text{s} t = 1 s
a ⃗ ( 1 ) = 6 ( 10 3 ) ( 1 ) i ^ + 2 ( 5 ) j ^ = 20 i ^ + 10 j ^ ms − 2 \vec{a}(1) = 6\left(\frac{10}{3}\right)(1)\hat{i} + 2(5)\hat{j} = 20\hat{i} + 10\hat{j}\ \text{ms}^{-2} a ( 1 ) = 6 ( 3 10 ) ( 1 ) i ^ + 2 ( 5 ) j ^ = 20 i ^ + 10 j ^ ms − 2 Calculate the force using Newton's second law F ⃗ = m a ⃗ \vec{F} = m\vec{a} F = m a t = 1 s t = 1\,\text{s} t = 1 s
F ⃗ ( 1 ) = 0.1 ( 20 i ^ + 10 j ^ ) \vec{F}(1) = 0.1(20\hat{i} + 10\hat{j}) F ( 1 ) = 0.1 ( 20 i ^ + 10 j ^ ) F ⃗ ( 1 ) = 2 i ^ + j ^ N \vec{F}(1) = 2\hat{i} + \hat{j}\ \text{N} F ( 1 ) = 2 i ^ + j ^ N Thus, Option (3) is FALSE .
Step 4: Find the angular momentum about the origin at t = 1 s t = 1\,\text{s} t = 1 s
The angular momentum L ⃗ \vec{L} L
L ⃗ ( t ) = m ( r ⃗ × v ⃗ ) \vec{L}(t) = m(\vec{r} \times \vec{v}) L ( t ) = m ( r × v ) Compute the cross product of r ⃗ ( t ) \vec{r}(t) r ( t ) v ⃗ ( t ) \vec{v}(t) v ( t )
r ⃗ × v ⃗ = ( α t 3 i ^ + β t 2 j ^ ) × ( 3 α t 2 i ^ + 2 β t j ^ ) \vec{r} \times \vec{v} = (\alpha t^{3}\hat{i} + \beta t^{2}\hat{j}) \times (3\alpha t^{2}\hat{i} + 2\beta t\hat{j}) r × v = ( α t 3 i ^ + β t 2 j ^ ) × ( 3 α t 2 i ^ + 2 β t j ^ ) r ⃗ × v ⃗ = ( 2 α β t 4 − 3 α β t 4 ) k ^ = − α β t 4 k ^ \vec{r} \times \vec{v} = (2\alpha\beta t^{4} - 3\alpha\beta t^{4})\hat{k} = -\alpha\beta t^{4}\hat{k} r × v = ( 2 α β t 4 − 3 α β t 4 ) k ^ = − α β t 4 k ^ Substitute this back into the angular momentum expression.
L ⃗ ( t ) = − m α β t 4 k ^ \vec{L}(t) = -m\alpha\beta t^{4}\hat{k} L ( t ) = − m α β t 4 k ^ Evaluate at t = 1 s t = 1\,\text{s} t = 1 s
L ⃗ ( 1 ) = − ( 0.1 ) ( 10 3 ) ( 5 ) ( 1 ) 4 k ^ \vec{L}(1) = -(0.1)\left(\frac{10}{3}\right)(5)(1)^{4}\hat{k} L ( 1 ) = − ( 0.1 ) ( 3 10 ) ( 5 ) ( 1 ) 4 k ^ L ⃗ ( 1 ) = − 5 3 k ^ N m s \vec{L}(1) = -\frac{5}{3}\hat{k}\ \text{N m s} L ( 1 ) = − 3 5 k ^ N m s Thus, Option (2) is TRUE .
Step 5: Find the torque about the origin at t = 1 s t = 1\,\text{s} t = 1 s
The torque τ ⃗ \vec{\tau} τ
τ ⃗ ( t ) = d L ⃗ d t = − 4 m α β t 3 k ^ \vec{\tau}(t) = \frac{d\vec{L}}{dt} = -4m\alpha\beta t^{3}\hat{k} τ ( t ) = d t d L = − 4 m α β t 3 k ^ Evaluate at t = 1 s t = 1\,\text{s} t = 1 s
τ ⃗ ( 1 ) = − 4 ( 0.1 ) ( 10 3 ) ( 5 ) ( 1 ) 3 k ^ \vec{\tau}(1) = -4(0.1)\left(\frac{10}{3}\right)(5)(1)^{3}\hat{k} τ ( 1 ) = − 4 ( 0.1 ) ( 3 10 ) ( 5 ) ( 1 ) 3 k ^ τ ⃗ ( 1 ) = − 20 3 k ^ N m \vec{\tau}(1) = -\frac{20}{3}\hat{k}\ \text{N m} τ ( 1 ) = − 3 20 k ^ N m Thus, Option (4) is TRUE .
Answer:
The true statements are (1), (2), and (4) :
v ⃗ ( 1 ) = 10 i ^ + 10 j ^ ms − 1 \vec{v}(1) = 10\hat{i} + 10\hat{j}\ \text{ms}^{-1} v ( 1 ) = 10 i ^ + 10 j ^ ms − 1 L ⃗ ( 1 ) = − 5 3 k ^ N m s \vec{L}(1) = -\dfrac{5}{3}\hat{k}\ \text{N m s} L ( 1 ) = − 3 5 k ^ N m s τ ⃗ ( 1 ) = − 20 3 k ^ N m \vec{\tau}(1) = -\dfrac{20}{3}\hat{k}\ \text{N m} τ ( 1 ) = − 3 20 k ^ N m 