Sample Solution

BPHCT-131 Solved Assignment

1. a) Determine the torque about the point $(0,1,1)$$(0,1,1)$(0,1,1)(0,1,1)$(0,1,1)$ due to a force $\overrightarrow{\mathbf{F}}=2\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$$\overrightarrow{\mathbf{F}}=2\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+\widehat{\mathbf{k}}$vec(F)=2 hat(i)- hat(j)+ hat(k)\overrightarrow{\mathbf{F}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$\overrightarrow{\mathbf{F}}=2\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ being exerted at the point $(4,2,3)$$(4,2,3)$(4,2,3)(4,2,3)$(4,2,3)$.
   b) Given two vector functions $\overrightarrow{\mathbf{a}}(t)=(t^3-t)\hat{\mathbf{i}}+(3t+4)\hat{\mathbf{j}}+2t^2\hat{\mathbf{k}}$$\overrightarrow{\mathbf{a}}(t)=\left(t^3-t\right)\widehat{\mathbf{i}}+(3t+4)\widehat{\mathbf{j}}+2t^2\widehat{\mathbf{k}}$vec(a)(t)=(t^(3)-t) hat(i)+(3t+4) hat(j)+2t^(2) hat(k)\overrightarrow{\mathbf{a}}(t)=\left(t^3-t\right) \hat{\mathbf{i}}+(3 t+4) \hat{\mathbf{j}}+2 t^2 \hat{\mathbf{k}}$\overrightarrow{\mathbf{a}}(t)=(t^3-t)\hat{\mathbf{i}}+(3t+4)\hat{\mathbf{j}}+2t^2\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}(t)=(7-t^2)\hat{\mathbf{i}}+(4+6t)\hat{\mathbf{j}}-\left(6t^3\right)\hat{\mathbf{k}}$$\overrightarrow{\mathbf{b}}(t)=\left(7-t^2\right)\widehat{\mathbf{i}}+(4+6t)\widehat{\mathbf{j}}-\left(6t^3\right)\widehat{\mathbf{k}}$vec(b)(t)=(7-t^(2)) hat(i)+(4+6t) hat(j)-(6t^(3)) hat(k)\overrightarrow{\mathbf{b}}(t)=\left(7-t^2\right) \hat{\mathbf{i}}+(4+6 t) \hat{\mathbf{j}}-\left(6 t^3\right) \hat{\mathbf{k}}$\overrightarrow{\mathbf{b}}(t)=(7-t^2)\hat{\mathbf{i}}+(4+6t)\hat{\mathbf{j}}-\left(6t^3\right)\hat{\mathbf{k}}$, determine the derivative of $\overrightarrow{\mathbf{a}}(t).\overrightarrow{\mathbf{b}}(t)$$\overrightarrow{\mathbf{a}}(t).\overrightarrow{\mathbf{b}}(t)$vec(a)(t). vec(b)(t)\overrightarrow{\mathbf{a}}(t) . \overrightarrow{\mathbf{b}}(t)$\overrightarrow{\mathbf{a}}(t).\overrightarrow{\mathbf{b}}(t)$ at $t=1$$t=1$t=1t=1$t=1$.
2. Solve the following ordinary differential equations:
   a) $(2y{x}^2+4)\frac{dy}{dx}+(2y^{2}x-3)=0$$\left(2y{x}^2+4\right)\frac{dy}{dx}+\left(2y^{2}x-3\right)=0$(2yx^(2)+4)(dy)/(dx)+(2y^(2)x-3)=0\left(2 y x^2+4\right) \frac{d y}{d x}+\left(2 y^2 x-3\right)=0$(2y{x}^2+4)\frac{dy}{dx}+(2y^{2}x-3)=0$
   b) $\frac{d^{2}y}{d{x}^2}-6\frac{dy}{dx}+13y=0$$\frac{d^{2}y}{d{x}^2}-6\frac{dy}{dx}+13y=0$(d^(2)y)/(dx^(2))-6(dy)/(dx)+13 y=0\frac{d^2 y}{d x^2}-6 \frac{d y}{d x}+13 y=0$\frac{d^{2}y}{d{x}^2}-6\frac{dy}{dx}+13y=0$ for $y(0)=2,y\left(\frac{\pi }{4}\right)=3$$y(0)=2,y\left(\frac{\pi }{4}\right)=3$y(0)=2,quad y((pi)/(4))=3y(0)=2, \quad y\left(\frac{\pi}{4}\right)=3$y(0)=2,y\left(\frac{\pi }{4}\right)=3$.
3. a) A box of mass 10 kg is being pulled on the floor by a mass-less rope with a force of 100 N at an angle of ${60}^{\circ }$${60}^{\circ }$60^(@)60^{\circ}${60}^{\circ }$ to the horizontal. What is the acceleration of the box if the coefficient of kinetic friction between the floor and the box is ${\mu }_k=0.25$${\mu }_k=0.25$mu _(k)=0.25\mu_k=0.25${\mu }_k=0.25$ ? Take $g=10{\mathrm{ms}}^{-2}$$g=10{\mathrm{ms}}^{-2}$g=10ms^(-2)g=10 \mathrm{~ms}^{-2}$g=10{\mathrm{ms}}^{-2}$.
   b) A ball having a mass of 0.5 kg is moving towards the east with a speed of velocity $6.0{\mathrm{ms}}^{-1}$$6.0{\mathrm{ms}}^{-1}$6.0ms^(-1)6.0 \mathrm{~ms}^{-1}$6.0{\mathrm{ms}}^{-1}$. After being hit by a bat it changes its direction and starts moving towards the north with a speed of $5.0{\mathrm{ms}}^{-1}$$5.0{\mathrm{ms}}^{-1}$5.0ms^(-1)5.0 \mathrm{~ms}^{-1}$5.0{\mathrm{ms}}^{-1}$. If the time of impact is 0.1 s , calculate the impulse and average force acting on the ball.
   c) A block of mass 5.0 kg starts from rest and slides down a surface which corresponds to a quarter of a circle of 3.0 m radius. (i) If the curved surface is smooth, find the speed at the bottom. (ii) If the speed at the bottom is $2.0{\mathrm{ms}}^{-1}$$2.0{\mathrm{ms}}^{-1}$2.0ms^(-1)2.0 \mathrm{~ms}^{-1}$2.0{\mathrm{ms}}^{-1}$, calculate the energy dissipated due to friction in the descent. (iii) After the block reaches the horizontal with a speed of $2.0{\mathrm{ms}}^{-1}$$2.0{\mathrm{ms}}^{-1}$2.0ms^(-1)2.0 \mathrm{~ms}^{-1}$2.0{\mathrm{ms}}^{-1}$ it slides to a stop in a distance of 1.5 m . calculate the frictional force acting on the horizontal surface. Take $g=10{\mathrm{ms}}^{-2}$$g=10{\mathrm{ms}}^{-2}$g=10ms^(-2)g=10 \mathrm{~ms}^{-2}$g=10{\mathrm{ms}}^{-2}$.
   d) A small satellite is in a circular orbit around a planet at a distance of $4.0\times {10}^8\mathrm{m}$$4.0\times {10}^8\mathrm{m}$4.0 xx10^(8)m4.0 \times 10^8 \mathrm{~m}$4.0\times {10}^8\mathrm{m}$ from the centre of the planet. The orbital speed of the satellite is $200{\mathrm{ms}}^{-1}$$200{\mathrm{ms}}^{-1}$200ms^(-1)200 \mathrm{~ms}^{-1}$200{\mathrm{ms}}^{-1}$. What is the mass of the planet?

Answer:

a) Determine the torque about the point $(0,1,1)$$(0,1,1)$(0,1,1)(0,1,1)$(0,1,1)$ due to a force $\overrightarrow{\mathbf{F}}=2\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$$\overrightarrow{\mathbf{F}}=2\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+\widehat{\mathbf{k}}$vec(F)=2 hat(i)- hat(j)+ hat(k)\overrightarrow{\mathbf{F}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$\overrightarrow{\mathbf{F}}=2\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ being exerted at the point $(4,2,3)$$(4,2,3)$(4,2,3)(4,2,3)$(4,2,3)$.

Answer:

• $\overrightarrow{r}$$\overrightarrow{r}$vec(r)\overrightarrow{r}$\overrightarrow{r}$ is the position vector from the point about which we’re calculating the torque to the point where the force is applied.
• $\overrightarrow{\mathbf{F}}$$\overrightarrow{\mathbf{F}}$vec(F)\overrightarrow{\mathbf{F}}$\overrightarrow{\mathbf{F}}$ is the force vector.

Step 1: Calculate the position vector $\overrightarrow{r}$$\overrightarrow{r}$vec(r)\overrightarrow{r}$\overrightarrow{r}$

Step 2: Write the force vector $\overrightarrow{\mathbf{F}}$$\overrightarrow{\mathbf{F}}$vec(F)\overrightarrow{\mathbf{F}}$\overrightarrow{\mathbf{F}}$

Step 3: Compute the cross product $\overrightarrow{r}\times \overrightarrow{\mathbf{F}}$$\overrightarrow{r}\times \overrightarrow{\mathbf{F}}$vec(r)xx vec(F)\overrightarrow{r} \times \overrightarrow{\mathbf{F}}$\overrightarrow{r}\times \overrightarrow{\mathbf{F}}$

• For $\hat{\mathbf{i}}$$\widehat{\mathbf{i}}$hat(i)\hat{\mathbf{i}}$\hat{\mathbf{i}}$:
  $$\left|\begin{array}{cc}1& 2\\ -1& 1\end{array}\right|=(1)(1)-(-1)(2)=1+2=3$$$$\left|\begin{array}{cc}1& 2\\ -1& 1\end{array}\right|=(1)(1)-(-1)(2)=1+2=3$$|[1,2],[-1,1]|=(1)(1)-(-1)(2)=1+2=3\begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} = (1)(1) – (-1)(2) = 1 + 2 = 3$$\left|\begin{array}{cc}1& 2\\ -1& 1\end{array}\right|=(1)(1)-(-1)(2)=1+2=3$$
• For $\hat{\mathbf{j}}$$\widehat{\mathbf{j}}$hat(j)\hat{\mathbf{j}}$\hat{\mathbf{j}}$:
  $$\left|\begin{array}{cc}4& 2\\ 2& 1\end{array}\right|=(4)(1)-(2)(2)=4-4=0$$$$\left|\begin{array}{cc}4& 2\\ 2& 1\end{array}\right|=(4)(1)-(2)(2)=4-4=0$$|[4,2],[2,1]|=(4)(1)-(2)(2)=4-4=0\begin{vmatrix} 4 & 2 \\ 2 & 1 \end{vmatrix} = (4)(1) – (2)(2) = 4 – 4 = 0$$\left|\begin{array}{cc}4& 2\\ 2& 1\end{array}\right|=(4)(1)-(2)(2)=4-4=0$$
• For $\hat{\mathbf{k}}$$\widehat{\mathbf{k}}$hat(k)\hat{\mathbf{k}}$\hat{\mathbf{k}}$:
  $$\left|\begin{array}{cc}4& 1\\ 2& -1\end{array}\right|=(4)(-1)-(1)(2)=-4-2=-6$$$$\left|\begin{array}{cc}4& 1\\ 2& -1\end{array}\right|=(4)(-1)-(1)(2)=-4-2=-6$$|[4,1],[2,-1]|=(4)(-1)-(1)(2)=-4-2=-6\begin{vmatrix} 4 & 1 \\ 2 & -1 \end{vmatrix} = (4)(-1) – (1)(2) = -4 – 2 = -6$$\left|\begin{array}{cc}4& 1\\ 2& -1\end{array}\right|=(4)(-1)-(1)(2)=-4-2=-6$$

Final Answer:

b) Given two vector functions $\overrightarrow{\mathbf{a}}(t)=(t^3-t)\hat{\mathbf{i}}+(3t+4)\hat{\mathbf{j}}+2t^2\hat{\mathbf{k}}$$\overrightarrow{\mathbf{a}}(t)=\left(t^3-t\right)\widehat{\mathbf{i}}+(3t+4)\widehat{\mathbf{j}}+2t^2\widehat{\mathbf{k}}$vec(a)(t)=(t^(3)-t) hat(i)+(3t+4) hat(j)+2t^(2) hat(k)\overrightarrow{\mathbf{a}}(t)=\left(t^3-t\right) \hat{\mathbf{i}}+(3 t+4) \hat{\mathbf{j}}+2 t^2 \hat{\mathbf{k}}$\overrightarrow{\mathbf{a}}(t)=(t^3-t)\hat{\mathbf{i}}+(3t+4)\hat{\mathbf{j}}+2t^2\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}(t)=(7-t^2)\hat{\mathbf{i}}+(4+6t)\hat{\mathbf{j}}-\left(6t^3\right)\hat{\mathbf{k}}$$\overrightarrow{\mathbf{b}}(t)=\left(7-t^2\right)\widehat{\mathbf{i}}+(4+6t)\widehat{\mathbf{j}}-\left(6t^3\right)\widehat{\mathbf{k}}$vec(b)(t)=(7-t^(2)) hat(i)+(4+6t) hat(j)-(6t^(3)) hat(k)\overrightarrow{\mathbf{b}}(t)=\left(7-t^2\right) \hat{\mathbf{i}}+(4+6 t) \hat{\mathbf{j}}-\left(6 t^3\right) \hat{\mathbf{k}}$\overrightarrow{\mathbf{b}}(t)=(7-t^2)\hat{\mathbf{i}}+(4+6t)\hat{\mathbf{j}}-\left(6t^3\right)\hat{\mathbf{k}}$, determine the derivative of $\overrightarrow{\mathbf{a}}(t).\overrightarrow{\mathbf{b}}(t)$$\overrightarrow{\mathbf{a}}(t).\overrightarrow{\mathbf{b}}(t)$vec(a)(t). vec(b)(t)\overrightarrow{\mathbf{a}}(t) . \overrightarrow{\mathbf{b}}(t)$\overrightarrow{\mathbf{a}}(t).\overrightarrow{\mathbf{b}}(t)$ at $t=1$$t=1$t=1t=1$t=1$.

Answer:

Step 1: Find the dot product $\overrightarrow{\mathbf{a}}(t)\cdot \overrightarrow{\mathbf{b}}(t)$$\overrightarrow{\mathbf{a}}(t)\cdot \overrightarrow{\mathbf{b}}(t)$vec(a)(t)* vec(b)(t)\overrightarrow{\mathbf{a}}(t) \cdot \overrightarrow{\mathbf{b}}(t)$\overrightarrow{\mathbf{a}}(t)\cdot \overrightarrow{\mathbf{b}}(t)$

• $a_1(t)$$a_1(t)$a_(1)(t)a_1(t)$a_1(t)$, $a_2(t)$$a_2(t)$a_(2)(t)a_2(t)$a_2(t)$, $a_3(t)$$a_3(t)$a_(3)(t)a_3(t)$a_3(t)$ are the components of $\overrightarrow{\mathbf{a}}(t)$$\overrightarrow{\mathbf{a}}(t)$vec(a)(t)\overrightarrow{\mathbf{a}}(t)$\overrightarrow{\mathbf{a}}(t)$
• $b_1(t)$$b_1(t)$b_(1)(t)b_1(t)$b_1(t)$, $b_2(t)$$b_2(t)$b_(2)(t)b_2(t)$b_2(t)$, $b_3(t)$$b_3(t)$b_(3)(t)b_3(t)$b_3(t)$ are the components of $\overrightarrow{\mathbf{b}}(t)$$\overrightarrow{\mathbf{b}}(t)$vec(b)(t)\overrightarrow{\mathbf{b}}(t)$\overrightarrow{\mathbf{b}}(t)$

• $\overrightarrow{\mathbf{a}}(t)=(t^3-t)\hat{\mathbf{i}}+(3t+4)\hat{\mathbf{j}}+2t^2\hat{\mathbf{k}}$$\overrightarrow{\mathbf{a}}(t)=(t^3-t)\widehat{\mathbf{i}}+(3t+4)\widehat{\mathbf{j}}+2t^2\widehat{\mathbf{k}}$vec(a)(t)=(t^(3)-t) hat(i)+(3t+4) hat(j)+2t^(2) hat(k)\overrightarrow{\mathbf{a}}(t) = (t^3 – t) \hat{\mathbf{i}} + (3t + 4) \hat{\mathbf{j}} + 2t^2 \hat{\mathbf{k}}$\overrightarrow{\mathbf{a}}(t)=(t^3-t)\hat{\mathbf{i}}+(3t+4)\hat{\mathbf{j}}+2t^2\hat{\mathbf{k}}$
• $\overrightarrow{\mathbf{b}}(t)=(7-t^2)\hat{\mathbf{i}}+(4+6t)\hat{\mathbf{j}}-6t^3\hat{\mathbf{k}}$$\overrightarrow{\mathbf{b}}(t)=(7-t^2)\widehat{\mathbf{i}}+(4+6t)\widehat{\mathbf{j}}-6t^3\widehat{\mathbf{k}}$vec(b)(t)=(7-t^(2)) hat(i)+(4+6t) hat(j)-6t^(3) hat(k)\overrightarrow{\mathbf{b}}(t) = (7 – t^2) \hat{\mathbf{i}} + (4 + 6t) \hat{\mathbf{j}} – 6t^3 \hat{\mathbf{k}}$\overrightarrow{\mathbf{b}}(t)=(7-t^2)\hat{\mathbf{i}}+(4+6t)\hat{\mathbf{j}}-6t^3\hat{\mathbf{k}}$

• $a_1(t)=t^3-t$$a_1(t)=t^3-t$a_(1)(t)=t^(3)-ta_1(t) = t^3 – t$a_1(t)=t^3-t$
• $a_2(t)=3t+4$$a_2(t)=3t+4$a_(2)(t)=3t+4a_2(t) = 3t + 4$a_2(t)=3t+4$
• $a_3(t)=2t^2$$a_3(t)=2t^2$a_(3)(t)=2t^(2)a_3(t) = 2t^2$a_3(t)=2t^2$
• $b_1(t)=7-t^2$$b_1(t)=7-t^2$b_(1)(t)=7-t^(2)b_1(t) = 7 – t^2$b_1(t)=7-t^2$
• $b_2(t)=4+6t$$b_2(t)=4+6t$b_(2)(t)=4+6tb_2(t) = 4 + 6t$b_2(t)=4+6t$
• $b_3(t)=-6t^3$$b_3(t)=-6t^3$b_(3)(t)=-6t^(3)b_3(t) = -6t^3$b_3(t)=-6t^3$

• First term: $(t^3-t)(7-t^2)$$(t^3-t)(7-t^2)$(t^(3)-t)(7-t^(2))(t^3 – t)(7 – t^2)$(t^3-t)(7-t^2)$
  $$=t^3(7-t^2)-t(7-t^2)=7t^3-t^5-7t+t^3=8t^3-t^5-7t$$$$=t^3(7-t^2)-t(7-t^2)=7t^3-t^5-7t+t^3=8t^3-t^5-7t$$=t^(3)(7-t^(2))-t(7-t^(2))=7t^(3)-t^(5)-7t+t^(3)=8t^(3)-t^(5)-7t= t^3(7 – t^2) – t(7 – t^2) = 7t^3 – t^5 – 7t + t^3 = 8t^3 – t^5 – 7t$$=t^3(7-t^2)-t(7-t^2)=7t^3-t^5-7t+t^3=8t^3-t^5-7t$$
• Second term: $(3t+4)(4+6t)$$(3t+4)(4+6t)$(3t+4)(4+6t)(3t + 4)(4 + 6t)$(3t+4)(4+6t)$
  $$=3t(4+6t)+4(4+6t)=12t+18t^2+16+24t=18t^2+36t+16$$$$=3t(4+6t)+4(4+6t)=12t+18t^2+16+24t=18t^2+36t+16$$=3t(4+6t)+4(4+6t)=12 t+18t^(2)+16+24 t=18t^(2)+36 t+16= 3t(4 + 6t) + 4(4 + 6t) = 12t + 18t^2 + 16 + 24t = 18t^2 + 36t + 16$$=3t(4+6t)+4(4+6t)=12t+18t^2+16+24t=18t^2+36t+16$$
• Third term: $(2t^2)(-6t^3)$$(2t^2)(-6t^3)$(2t^(2))(-6t^(3))(2t^2)(-6t^3)$(2t^2)(-6t^3)$
  $$=-12t^5$$$$=-12t^5$$=-12t^(5)= -12t^5$$=-12t^5$$

Step 2: Differentiate the dot product

Step 3: Evaluate the derivative at $t=1$$t=1$t=1t = 1$t=1$

Final Answer: