Q: A vehicle is moving with a speed \$v\$ on a curved smooth road of width \$b\$ and radius \$R\$. For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is

tan \$\theta \$ \$=\dfrac{v^{2}}{Rg} \Rightarrow \dfrac{h}{b}=\dfrac{v^{2}}{Rg} \Rightarrow h=\dfrac{v^{2}b}{Rg}\$.

Q: A uniform ladder of length \$5 m\$ is placed against the wall as shown in the figure. If coefficient of friction \$\mu \$ is the same for both the walls, what is the minimum value of \$\mu \$ for it not to slip ?

\$\textbf{Hint}\$: Draw the FBD \$\textbf{Step 1}\$:Given length of ladder is 5m. It is placed against the wall. Coefficient of friction is \$\mu\$. Let mass of the ladder is \$m\$.From the figure: for the vertical equilibrium \$N_1+\mu N_2=mg\$ \$(1)\$For horizontal equilibrium \$N_2=\mu N_1\$ \$(2)\$The torque about point A is given by: \$mg(2.5)cos \theta=N_2(4)+\mu N_2(3)\$\$mg(2.5 \times \dfrac{3}{5})=(4+3 \mu)N_2\$, since \$cos \theta=\dfrac{3}{5}\$\$mg(\dfrac{3}{2})=(4+3 \mu)N_2\$ \$(3)\$\textbf{Step 2}\$:From equation \$(1)\$ and \$(2)\$ \$N_2(\dfrac{1+\mu^2}{\mu})=mg \implies N_2= \dfrac{\mu mg}{1+\mu^2}\$So, \$mg \dfrac{3}{2}=(4+3 \mu) \dfrac{\mu mg}{1+\mu^2}\$\$(3+3 \mu^2)=8 \mu+6 \mu^2\$\$3 \mu^2+8 \mu-3=0\$\$\mu=\dfrac{-8 \pm sqrt{64+36}}{6}=\dfrac{-8 \pm 10}{6}\$\$\mu=\dfrac{1}{3} Thus option C is correct.

Q: The horizontal component \$N_{x}\$ and vertical component \$N_{y}\$ of the reaction at end P are given by:

For equilibrium,\$\sum F_{x}=0\$ thus\$Tcos37^{0}-N_{x}=0\$\$ N_{x}=\displaystyle\frac{4}{5}T\$\$=\displaystyle\frac{4}{5}(1200)\$\$=960N\$\$\sum F_{y}=0\$, thus\$N_{y}-Tsin37^{0}+500+80=0\$or\$N_{y}=(1200)\left( \displaystyle\frac{3}{5} \right )-500-80\$\$=140N\$

Q: Two persons \$P\$ and \$Q\$ of the same height are carrying a uniform beam of length \$3 m\$. If \$Q\$ is at one end, then the distance of \$P\$ from the other end such that \$P\$, \$Q\$ receive loads in the ratio \$5 : 3\$ is:

\$N_{P}+N_{Q}=Mg\$also \$\dfrac{N_{P}}{N_{Q}}=\dfrac{5}{3}\$\$ \Rightarrow N_{P}=\dfrac{5}{8}Mg; N_{Q}=\dfrac{3}{8}Mg\$Calculating torque about the end opposite to \$Q\$\$N_{P}d_{1}+N_{Q}(3-d_{2})=Mg\dfrac{3}{2}\$\$\dfrac{5}{8}Mgd_{1}+\dfrac{3}{8}Mg \times 3=Mg\dfrac{3}{2}\$\$\therefore \dfrac{5}{8}Mgd_{1}+\dfrac{3}{8}Mg\times 3=Mg\dfrac{3}{2}\$\$\dfrac{5}{8}Mgd_{1}=(\dfrac{12-9}{8})Mg\$\$d_{1}=\dfrac{3}{5}=0.6\: m\$

Q: The tension in the string is:

For equilibrium about P \$\sum \tau _{p}=0\$or\$[(500)1+80(0.5)-(Tsin37^{0})(0.75)]=0\$(note that torque produced by \$Tcos30^{0}\$, \$N_{x}\$ and \$N_{y}\$ about the point A is equal to zero),Thus\$500+40-T\left ( \displaystyle\frac{3}{5} \right )(0.75)=0\$or\$T=\displaystyle\frac{540(5)}{3(0.75)}=1200N\$

Q: A stationary ball weighing \$0.25\ kg\$ acquires a speed of \$10 \ m/s\$ when hit by a hockey stick. The impulse imparted to the ball is:

Given that the mass of the body is \$0.25 kg\$ is at rest initially. when the body is hit by the stick it gains speed of \$10\ m/s\$.means initial speed \$u = 0\$ and after getting hit it's final speed \$ v = 10 m/s\$.we know that impulse is equal to change in momentum i.e. \$\text {Impulse } J =m(v-u)\$so here \$ J = 0.25 ( 10 - 0 )\$ \$kg\ m/s\$ or \$ N \$hence The impulse imparted to the ball is \$2.5 N s\$

Q: An elastic string carrying a body of mass \$m\$ at one end extends by \$1.5\ cm\$. If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

When the body is at rest, the extension is \$1.5\ cm\$ while force acting on it is \$ mg \$Force when it reaches bottom is \$F_b = \dfrac{mv^2}{r} + mg \$Now using energy balance between topmost point and bottom most point ,\$\dfrac{1}{2}mv^2=\dfrac{1}{2}m(\sqrt{rg})^2+mgh\$ (h is the height difference between the top and bottom point ; \$\sqrt{gr}\$ is critical velocity at topmost point )\$ v^2 = rg + 2gh\$or, \$ v^2 = rg + 4gr = 5 gr\$So, \$F_b = 6mg \$Using this in: \$\dfrac{e_2}{e_1} = \dfrac{F_2}{F_1} \$, we get:\$e_2 = 1.5 \times 6 = 9.0\: cm \$

Q: Two identical ladders are arranged as shown in the figure. Mass of each ladder is \$M\$ and length \$L\$. The system is in equilibrium. Find direction and magnitude of friction force acting at A or B.

Drawing the FBD of both rods As the rods are in equilibrium\$\Sigma F_{X}=0\$\$\Sigma F_{y}=0\$\$\tau_{\mathrm{n}\mathrm{e}\mathrm{t}}=0\$For right rod\$N_{2}+\displaystyle \frac{mg}{2}+Mg=N\$ (1)\$N_{1}=f\$ (2)For left rod\$N_{2}+N=Mg+\displaystyle \frac{mg}{2}\$ (3)From \$(1)\mathrm{a}\mathrm{n}\mathrm{d}(3)N_{1}=f\$\$= N_{2}=0 = N=Mg+\displaystyle \frac{mg}{2}\$Balancing torque about \$\mathrm{O}\$,\$ Mg\displaystyle \frac{L}{2}\cos\theta +fL \sin\theta =NL\cos\theta\$\$\displaystyle \frac{Mg\cos\theta}{2}+f\sin\theta =(Mg+\displaystyle \frac{mg}{2})\cos\theta\$\$ f=(\displaystyle \frac{Mg}{2}+\frac{mg}{2})\cot\theta = f=(\displaystyle \frac{M+m}{2})g\cot\theta\$

Question 1

Which of the following statements for a rigid object undergoing pure translational motion are false?

Accepted Answer

(A) For head on elastic collision of sphere with wall, K.E. of sphere does not change but the sphere receives an impulse. Hence false.(B) If K.E. of a sphere changes its velocity and thus momentum must change. Hence false.(C) A force acting perpendicular to velocity changes momentum, but does no work. Hence true.(D) When a force does work, kinetic energy of body changes. If K.E. changes, velocity and thus momentum must change. Hence false.

Question 2

A very long uniform helix is made of thin metal wire. The axis of helix is vertical. A small bead begins to slide down the fixed helix starting from rest. Considering friction between bead and wire of helix to be non-zero, which of the following statements is/ are true as long as bead moves on helix.

Accepted Answer

hint: first draw FBD of system and then applied Newton second lawstep-1: draw FBD of systemstep-2: apply Newton second lawstep-3: conclusion of motion

Question 3

A vehicle is moving with a speed $v$ on a curved smooth road of width $b$ and radius $R$ . For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is

Accepted Answer

tan $\theta $ $=\dfrac{v^{2}}{Rg} \Rightarrow \dfrac{h}{b}=\dfrac{v^{2}}{Rg} \Rightarrow h=\dfrac{v^{2}b}{Rg}$.

Question 4

A uniform ladder of length $5 m$ is placed against the wall as shown in the figure. If coefficient of friction $μ$ is the same for both the walls, what is the minimum value of $μ$ for it not to slip ?

Accepted Answer

$\textbf{Hint}$: Draw the FBD $\textbf{Step 1}$:Given length of ladder is 5m. It is placed against the wall. Coefficient of friction is $\mu$. Let mass of the ladder is $m$.From the figure: for the vertical equilibrium $N_1+\mu N_2=mg$  $(1)$For horizontal equilibrium $N_2=\mu N_1$     $(2)$The torque about point A is given by: $mg(2.5)cos \theta=N_2(4)+\mu N_2(3)$$mg(2.5 \times \dfrac{3}{5})=(4+3 \mu)N_2$, since $cos \theta=\dfrac{3}{5}$$mg(\dfrac{3}{2})=(4+3 \mu)N_2$    $(3)$\textbf{Step 2}$:From equation $(1)$ and $(2)N_2(\dfrac{1+\mu^2}{\mu})=mg \implies N_2= \dfrac{\mu mg}{1+\mu^2}$So, $mg \dfrac{3}{2}=(4+3 \mu) \dfrac{\mu mg}{1+\mu^2}(3+3 \mu^2)=8 \mu+6 \mu^23 \mu^2+8 \mu-3=0\mu=\dfrac{-8 \pm sqrt{64+36}}{6}=\dfrac{-8 \pm 10}{6}\mu=\dfrac{1}{3} Thus option C is correct.

Question 5

The horizontal component $N_{x}$ and vertical component $N_{y}$ of the reaction at end P are given by:

Accepted Answer

For equilibrium,$\sum F_{x}=0$ thus$Tcos37^{0}-N_{x}=0$$ N_{x}=\displaystyle\frac{4}{5}T$$=\displaystyle\frac{4}{5}(1200)$$=960N$$\sum F_{y}=0$, thus$N_{y}-Tsin37^{0}+500+80=0$or$N_{y}=(1200)\left( \displaystyle\frac{3}{5} \right )-500-80$$=140N$

Question 6

Two persons $P$ and $Q$ of the same height are carrying a uniform beam of length $3 m$ . If $Q$ is at one end, then the distance of $P$ from the other end such that $P$ , $Q$ receive loads in the ratio $5 : 3$ is:

Accepted Answer

$N_{P}+N_{Q}=Mg$also $\dfrac{N_{P}}{N_{Q}}=\dfrac{5}{3}$$ \Rightarrow N_{P}=\dfrac{5}{8}Mg; N_{Q}=\dfrac{3}{8}Mg$Calculating torque about the end opposite to $Q$$N_{P}d_{1}+N_{Q}(3-d_{2})=Mg\dfrac{3}{2}$$\dfrac{5}{8}Mgd_{1}+\dfrac{3}{8}Mg \times 3=Mg\dfrac{3}{2}$$\therefore \dfrac{5}{8}Mgd_{1}+\dfrac{3}{8}Mg\times 3=Mg\dfrac{3}{2}$$\dfrac{5}{8}Mgd_{1}=(\dfrac{12-9}{8})Mg$$d_{1}=\dfrac{3}{5}=0.6\: m$

Question 7

The tension in the string is:

Accepted Answer

For equilibrium about P $\sum \tau _{p}=0$or$[(500)1+80(0.5)-(Tsin37^{0})(0.75)]=0$(note that torque produced by $Tcos30^{0}$, $N_{x}$ and $N_{y}$ about the point A is equal to zero),Thus$500+40-T\left ( \displaystyle\frac{3}{5} \right )(0.75)=0$or$T=\displaystyle\frac{540(5)}{3(0.75)}=1200N$

Question 8

A stationary ball weighing $0.25\  kg$ acquires a speed of $10 \ m/s$ when hit by a hockey stick. The impulse imparted to the ball is:

Accepted Answer

Given that the mass of the body is $0.25 kg$ is at rest initially. when the body is hit by the stick it gains speed of  $10\ m/s$.means initial speed $u = 0$ and after getting hit it's final speed $ v = 10 m/s$.we know that impulse is equal to change in momentum i.e.  $\text {Impulse } J =m(v-u)$so here $ J = 0.25 ( 10 - 0 )$  $kg\ m/s$  or $ N $hence The impulse imparted to the ball is $2.5 N s$

Question 9

An elastic string carrying a body of mass $m$ at one end extends by $1.5\ cm$. If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

Accepted Answer

When the body is at rest, the extension is $1.5\ cm$ while force acting on it is $ mg $Force when it reaches bottom is $F_b = \dfrac{mv^2}{r} + mg $Now using energy balance between topmost point and bottom most point ,$\dfrac{1}{2}mv^2=\dfrac{1}{2}m(\sqrt{rg})^2+mgh$ (h is the height difference between the top and bottom point ; $\sqrt{gr}$ is critical velocity at topmost point )$ v^2 = rg + 2gh$or, $ v^2 = rg + 4gr = 5 gr$So, $F_b = 6mg $Using this in: $\dfrac{e_2}{e_1} = \dfrac{F_2}{F_1} $, we get:$e_2 = 1.5 \times 6 = 9.0\: cm $

Question 10

Two identical ladders are arranged as shown in the figure. Mass of each ladder is $M$ and length $L$. The system is in equilibrium. Find direction and magnitude of friction force acting at A or B.

Accepted Answer

Drawing the FBD of both rods As the rods are in equilibrium$\Sigma F_{X}=0$$\Sigma F_{y}=0$$\tau_{\mathrm{n}\mathrm{e}\mathrm{t}}=0$For right rod$N_{2}+\displaystyle \frac{mg}{2}+Mg=N$ (1)$N_{1}=f$ (2)For left rod$N_{2}+N=Mg+\displaystyle \frac{mg}{2}$ (3)From $(1)\mathrm{a}\mathrm{n}\mathrm{d}(3)N_{1}=f$$= N_{2}=0 = N=Mg+\displaystyle \frac{mg}{2}$Balancing torque about $\mathrm{O}$,$ Mg\displaystyle \frac{L}{2}\cos\theta +fL \sin\theta =NL\cos\theta$$\displaystyle \frac{Mg\cos\theta}{2}+f\sin\theta =(Mg+\displaystyle \frac{mg}{2})\cos\theta$$ f=(\displaystyle \frac{Mg}{2}+\frac{mg}{2})\cot\theta = f=(\displaystyle \frac{M+m}{2})g\cot\theta$

Laws Of Motion

Physics
13386 Views

Which of the following statements for a rigid object undergoing pure translational motion are false?

A vehicle is moving with a speed $v$ on a curved smooth road of width $b$ and radius $R$ . For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is

A uniform ladder of length $5 m$ is placed against the wall as shown in the figure. If coefficient of friction $μ$ is the same for both the walls, what is the minimum value of $μ$ for it not to slip ?

The horizontal component $N_{x}$ and vertical component $N_{y}$ of the reaction at end P are given by:

Two persons $P$ and $Q$ of the same height are carrying a uniform beam of length $3 m$ . If $Q$ is at one end, then the distance of $P$ from the other end such that $P$ , $Q$ receive loads in the ratio $5 : 3$ is:

The tension in the string is:

A stationary ball weighing $0.25 k g$ acquires a speed of $10 m / s$ when hit by a hockey stick. The impulse imparted to the ball is:

An elastic string carrying a body of mass $m$ at one end extends by $1.5 c m$ . If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

Two identical ladders are arranged as shown in the figure. Mass of each ladder is $M$ and length $L$ . The system is in equilibrium. Find direction and magnitude of friction force acting at A or B.

Laws Of Motion

Physics 13386 Views

Which of the following statements for a rigid object undergoing pure translational motion are false?

A vehicle is moving with a speed v on a curved smooth road of width b and radius R. For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is

A uniform ladder of length 5m is placed against the wall as shown in the figure. If coefficient of friction μ is the same for both the walls, what is the minimum value of μ for it not to slip ?

The horizontal component Nx​ and vertical component Ny​ of the reaction at end P are given by:

Two persons P and Q of the same height are carrying a uniform beam of length 3m. If Q is at one end, then the distance of P from the other end such that P, Q receive loads in the ratio 5:3 is:

The tension in the string is:

A stationary ball weighing 0.25 kg acquires a speed of 10 m/s when hit by a hockey stick. The impulse imparted to the ball is:

An elastic string carrying a body of mass m at one end extends by 1.5 cm. If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

Two identical ladders are arranged as shown in the figure. Mass of each ladder is M and length L. The system is in equilibrium. Find direction and magnitude of friction force acting at A or B.

Physics
13386 Views

A vehicle is moving with a speed $v$ on a curved smooth road of width $b$ and radius $R$ . For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is

A uniform ladder of length $5 m$ is placed against the wall as shown in the figure. If coefficient of friction $μ$ is the same for both the walls, what is the minimum value of $μ$ for it not to slip ?

The horizontal component $N_{x}$ and vertical component $N_{y}$ of the reaction at end P are given by:

Two persons $P$ and $Q$ of the same height are carrying a uniform beam of length $3 m$ . If $Q$ is at one end, then the distance of $P$ from the other end such that $P$ , $Q$ receive loads in the ratio $5 : 3$ is:

A stationary ball weighing $0.25 k g$ acquires a speed of $10 m / s$ when hit by a hockey stick. The impulse imparted to the ball is:

An elastic string carrying a body of mass $m$ at one end extends by $1.5 c m$ . If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

Two identical ladders are arranged as shown in the figure. Mass of each ladder is $M$ and length $L$ . The system is in equilibrium. Find direction and magnitude of friction force acting at A or B.