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Laws of Motion

146120 A solid cylinder of mass ' $m$ ' and radius ' $r$ ' starts rolling down an inclined plane of inclination $θ$ . If the friction is just enough to prevent slipping, the speed of its centre of mass after it has descended through a height ' $h$ ' is given by

1

\sqrt{\frac{4}{3}} gh

2

\sqrt{\frac{4 gh}{3}}

3

\frac{3}{4} gh

4

\sqrt{\frac{3 g h}{4}}

Explanation:

B For pure rolling, total mechanical energy will be conserved.
Friction work $= 0$ (zero)
From figure -

Energy at position $1, (E_{1}) = mgh$
Energy at position 2,
$E_{2} = \frac{1}{2} {mv}_{cm}^{2} + \frac{1}{2} I_{cm} ω^{2}$
$∵ \frac{v_{cm}}{r} = ω and I_{cm} = \frac{{mr}^{2}}{2}$
$E_{2} = \frac{1}{2} m \times r^{2} ω^{2} + \frac{1}{2} \frac{{mr}^{2}}{2} ω^{2}$
$E_{2} = \frac{1}{2} {mr}^{2} ω^{2} + \frac{1}{4} {mr}^{2} ω^{2}$
$E_{2} = \frac{3}{4} {mr}^{2} ω^{2}$
$E_{2} = \frac{3}{4} {mv}_{cm}^{2} (∵ v_{cm} = r ω)$
From law of conservation of energy $E_{1} = E_{2}$
$mgh = \frac{3}{4} {mv}_{cm}^{2}$
$v_{cm}^{2} = \frac{4}{3} gh$
$v_{cm} = \sqrt{\frac{4}{3} gh}$

AP EAMCET-03.09.2021

Laws of Motion

146121 A motor cyclist wants to drive in horizontal circles on the vertical inner surface of a large cylindrical wooden well of radius $8.0 m$ with minimum speed of $5 \sqrt{5} m \cdot s^{- 1}$ . The minimum value of coefficient of friction between the tires and the wall of the well must be - $(g = 10$ $m \cdot s^{- 2}$ )

1 0.10

2 0.64

3 0.30

4 0.40

Explanation:

B Given that-
Radius of wooden well $= 8.0 m$
Minimum speed $= 5 \sqrt{5} m \cdot s^{- 1}$
Friction force, $f_{s} = mg$
Centripetal force, $N = \frac{{mv}^{2}}{r}$
We know,
$μ = \frac{f_{s}}{N}$
$∴ μ = \frac{mg}{\frac{{mv}^{2}}{r}}$
$μ = \frac{r g}{v^{2}}$
$μ = \frac{8 \times 10}{(5 \sqrt{5})^{2}}$
$μ = \frac{80}{125}$
$μ = 0.64$

AP EAMCET-19.08.2021

Laws of Motion

146122 A block B, lying on a table, weighs ' $W$ '. The coefficient of static friction between the block and the table is $μ$ . Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

1

\frac{W \tan θ}{μ}

2

μ W \tan θ

3

μ W \sqrt{1 + \tan^{2} θ}

4

μ W \sin θ

Explanation:

B
The block B is under equilibrium by action of tension force ' $T$ '
Hence $T = μ N$ and $N = W$
So, $T = μ N = μ W$
At knot applying lami's theorem
$\frac{T}{\sin (90^{\circ} + θ)} = \frac{W_{A}}{\sin (180^{\circ} - θ)} = \frac{P}{\sin 90^{\circ}}$
$\frac{T}{\cos θ} = \frac{W_{A}}{\sin θ} = \frac{P}{1}$
$∴ W_{A} = \frac{T \sin θ}{\cos θ} = T \tan θ$
$W_{A} = μ W \tan θ {∵ T = μ W}$

AP EAMCET-23.08.2021

Laws of Motion

146123 A solid flywheel of mass $20 kg$ and radius 120 $mm$ revolves at $600 rpm$ . Find the total force that must applied by the brake so that the flywheel stop in 3 seconds. Given the coefficient of friction between the wheels and brake lining is 0.1 .

1

80 π N

2

24 π N

3

70 π N

4

60 π N

Explanation:

A Given that,
Speed $n = 600 rpm$
Moment of inertia $I = \frac{1}{2} {mR}^{2}$
Radius $R = 120 mm = 0.12 m$
Mass $m = 20 kg$
Coefficient of friction $μ = 0.1$
Angular speed $(ω) = \frac{2 π n}{60} = \frac{2 \times π \times 600}{60} = 20 π red / \sec$
Angular acceleration $(α) = \frac{Δ ω}{Δ t} = \frac{ω}{t} = \frac{20 π}{3} = \frac{20}{3} π red / s^{2}$
Further, torque $τ = I α$
Here, $μ F . R = I α$
$F = \frac{I α}{μ R}$
$F = \frac{\frac{1}{2} m^{2} \times \frac{20 π}{3}}{μ R}$
Or force, $F = \frac{10 π mR}{3 μ}$
Substituting the value, we get-
$F = \frac{10 π \times 20 \times 0.12}{3 \times 0.1}$
$F = 80 π N$

AP EAMCET (Medical)-05.10.2021

Laws of Motion

146120 A solid cylinder of mass ' $m$ ' and radius ' $r$ ' starts rolling down an inclined plane of inclination $θ$ . If the friction is just enough to prevent slipping, the speed of its centre of mass after it has descended through a height ' $h$ ' is given by

1

\sqrt{\frac{4}{3}} gh

2

\sqrt{\frac{4 gh}{3}}

3

\frac{3}{4} gh

4

\sqrt{\frac{3 g h}{4}}

Explanation:

B For pure rolling, total mechanical energy will be conserved.
Friction work $= 0$ (zero)
From figure -

Energy at position $1, (E_{1}) = mgh$
Energy at position 2,
$E_{2} = \frac{1}{2} {mv}_{cm}^{2} + \frac{1}{2} I_{cm} ω^{2}$
$∵ \frac{v_{cm}}{r} = ω and I_{cm} = \frac{{mr}^{2}}{2}$
$E_{2} = \frac{1}{2} m \times r^{2} ω^{2} + \frac{1}{2} \frac{{mr}^{2}}{2} ω^{2}$
$E_{2} = \frac{1}{2} {mr}^{2} ω^{2} + \frac{1}{4} {mr}^{2} ω^{2}$
$E_{2} = \frac{3}{4} {mr}^{2} ω^{2}$
$E_{2} = \frac{3}{4} {mv}_{cm}^{2} (∵ v_{cm} = r ω)$
From law of conservation of energy $E_{1} = E_{2}$
$mgh = \frac{3}{4} {mv}_{cm}^{2}$
$v_{cm}^{2} = \frac{4}{3} gh$
$v_{cm} = \sqrt{\frac{4}{3} gh}$

AP EAMCET-03.09.2021

Laws of Motion

146121 A motor cyclist wants to drive in horizontal circles on the vertical inner surface of a large cylindrical wooden well of radius $8.0 m$ with minimum speed of $5 \sqrt{5} m \cdot s^{- 1}$ . The minimum value of coefficient of friction between the tires and the wall of the well must be - $(g = 10$ $m \cdot s^{- 2}$ )

1 0.10

2 0.64

3 0.30

4 0.40

Explanation:

B Given that-
Radius of wooden well $= 8.0 m$
Minimum speed $= 5 \sqrt{5} m \cdot s^{- 1}$
Friction force, $f_{s} = mg$
Centripetal force, $N = \frac{{mv}^{2}}{r}$
We know,
$μ = \frac{f_{s}}{N}$
$∴ μ = \frac{mg}{\frac{{mv}^{2}}{r}}$
$μ = \frac{r g}{v^{2}}$
$μ = \frac{8 \times 10}{(5 \sqrt{5})^{2}}$
$μ = \frac{80}{125}$
$μ = 0.64$

AP EAMCET-19.08.2021

Laws of Motion

146122 A block B, lying on a table, weighs ' $W$ '. The coefficient of static friction between the block and the table is $μ$ . Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

1

\frac{W \tan θ}{μ}

2

μ W \tan θ

3

μ W \sqrt{1 + \tan^{2} θ}

4

μ W \sin θ

Explanation:

B
The block B is under equilibrium by action of tension force ' $T$ '
Hence $T = μ N$ and $N = W$
So, $T = μ N = μ W$
At knot applying lami's theorem
$\frac{T}{\sin (90^{\circ} + θ)} = \frac{W_{A}}{\sin (180^{\circ} - θ)} = \frac{P}{\sin 90^{\circ}}$
$\frac{T}{\cos θ} = \frac{W_{A}}{\sin θ} = \frac{P}{1}$
$∴ W_{A} = \frac{T \sin θ}{\cos θ} = T \tan θ$
$W_{A} = μ W \tan θ {∵ T = μ W}$

AP EAMCET-23.08.2021

Laws of Motion

146123 A solid flywheel of mass $20 kg$ and radius 120 $mm$ revolves at $600 rpm$ . Find the total force that must applied by the brake so that the flywheel stop in 3 seconds. Given the coefficient of friction between the wheels and brake lining is 0.1 .

1

80 π N

2

24 π N

3

70 π N

4

60 π N

Explanation:

A Given that,
Speed $n = 600 rpm$
Moment of inertia $I = \frac{1}{2} {mR}^{2}$
Radius $R = 120 mm = 0.12 m$
Mass $m = 20 kg$
Coefficient of friction $μ = 0.1$
Angular speed $(ω) = \frac{2 π n}{60} = \frac{2 \times π \times 600}{60} = 20 π red / \sec$
Angular acceleration $(α) = \frac{Δ ω}{Δ t} = \frac{ω}{t} = \frac{20 π}{3} = \frac{20}{3} π red / s^{2}$
Further, torque $τ = I α$
Here, $μ F . R = I α$
$F = \frac{I α}{μ R}$
$F = \frac{\frac{1}{2} m^{2} \times \frac{20 π}{3}}{μ R}$
Or force, $F = \frac{10 π mR}{3 μ}$
Substituting the value, we get-
$F = \frac{10 π \times 20 \times 0.12}{3 \times 0.1}$
$F = 80 π N$

AP EAMCET (Medical)-05.10.2021

Laws of Motion

146120 A solid cylinder of mass ' $m$ ' and radius ' $r$ ' starts rolling down an inclined plane of inclination $θ$ . If the friction is just enough to prevent slipping, the speed of its centre of mass after it has descended through a height ' $h$ ' is given by

1

\sqrt{\frac{4}{3}} gh

2

\sqrt{\frac{4 gh}{3}}

3

\frac{3}{4} gh

4

\sqrt{\frac{3 g h}{4}}

Explanation:

B For pure rolling, total mechanical energy will be conserved.
Friction work $= 0$ (zero)
From figure -

Energy at position $1, (E_{1}) = mgh$
Energy at position 2,
$E_{2} = \frac{1}{2} {mv}_{cm}^{2} + \frac{1}{2} I_{cm} ω^{2}$
$∵ \frac{v_{cm}}{r} = ω and I_{cm} = \frac{{mr}^{2}}{2}$
$E_{2} = \frac{1}{2} m \times r^{2} ω^{2} + \frac{1}{2} \frac{{mr}^{2}}{2} ω^{2}$
$E_{2} = \frac{1}{2} {mr}^{2} ω^{2} + \frac{1}{4} {mr}^{2} ω^{2}$
$E_{2} = \frac{3}{4} {mr}^{2} ω^{2}$
$E_{2} = \frac{3}{4} {mv}_{cm}^{2} (∵ v_{cm} = r ω)$
From law of conservation of energy $E_{1} = E_{2}$
$mgh = \frac{3}{4} {mv}_{cm}^{2}$
$v_{cm}^{2} = \frac{4}{3} gh$
$v_{cm} = \sqrt{\frac{4}{3} gh}$

AP EAMCET-03.09.2021

Laws of Motion

146121 A motor cyclist wants to drive in horizontal circles on the vertical inner surface of a large cylindrical wooden well of radius $8.0 m$ with minimum speed of $5 \sqrt{5} m \cdot s^{- 1}$ . The minimum value of coefficient of friction between the tires and the wall of the well must be - $(g = 10$ $m \cdot s^{- 2}$ )

1 0.10

2 0.64

3 0.30

4 0.40

Explanation:

B Given that-
Radius of wooden well $= 8.0 m$
Minimum speed $= 5 \sqrt{5} m \cdot s^{- 1}$
Friction force, $f_{s} = mg$
Centripetal force, $N = \frac{{mv}^{2}}{r}$
We know,
$μ = \frac{f_{s}}{N}$
$∴ μ = \frac{mg}{\frac{{mv}^{2}}{r}}$
$μ = \frac{r g}{v^{2}}$
$μ = \frac{8 \times 10}{(5 \sqrt{5})^{2}}$
$μ = \frac{80}{125}$
$μ = 0.64$

AP EAMCET-19.08.2021

Laws of Motion

146122 A block B, lying on a table, weighs ' $W$ '. The coefficient of static friction between the block and the table is $μ$ . Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

1

\frac{W \tan θ}{μ}

2

μ W \tan θ

3

μ W \sqrt{1 + \tan^{2} θ}

4

μ W \sin θ

Explanation:

B
The block B is under equilibrium by action of tension force ' $T$ '
Hence $T = μ N$ and $N = W$
So, $T = μ N = μ W$
At knot applying lami's theorem
$\frac{T}{\sin (90^{\circ} + θ)} = \frac{W_{A}}{\sin (180^{\circ} - θ)} = \frac{P}{\sin 90^{\circ}}$
$\frac{T}{\cos θ} = \frac{W_{A}}{\sin θ} = \frac{P}{1}$
$∴ W_{A} = \frac{T \sin θ}{\cos θ} = T \tan θ$
$W_{A} = μ W \tan θ {∵ T = μ W}$

AP EAMCET-23.08.2021

Laws of Motion

146123 A solid flywheel of mass $20 kg$ and radius 120 $mm$ revolves at $600 rpm$ . Find the total force that must applied by the brake so that the flywheel stop in 3 seconds. Given the coefficient of friction between the wheels and brake lining is 0.1 .

1

80 π N

2

24 π N

3

70 π N

4

60 π N

Explanation:

A Given that,
Speed $n = 600 rpm$
Moment of inertia $I = \frac{1}{2} {mR}^{2}$
Radius $R = 120 mm = 0.12 m$
Mass $m = 20 kg$
Coefficient of friction $μ = 0.1$
Angular speed $(ω) = \frac{2 π n}{60} = \frac{2 \times π \times 600}{60} = 20 π red / \sec$
Angular acceleration $(α) = \frac{Δ ω}{Δ t} = \frac{ω}{t} = \frac{20 π}{3} = \frac{20}{3} π red / s^{2}$
Further, torque $τ = I α$
Here, $μ F . R = I α$
$F = \frac{I α}{μ R}$
$F = \frac{\frac{1}{2} m^{2} \times \frac{20 π}{3}}{μ R}$
Or force, $F = \frac{10 π mR}{3 μ}$
Substituting the value, we get-
$F = \frac{10 π \times 20 \times 0.12}{3 \times 0.1}$
$F = 80 π N$

AP EAMCET (Medical)-05.10.2021

Laws of Motion

146120 A solid cylinder of mass ' $m$ ' and radius ' $r$ ' starts rolling down an inclined plane of inclination $θ$ . If the friction is just enough to prevent slipping, the speed of its centre of mass after it has descended through a height ' $h$ ' is given by

1

\sqrt{\frac{4}{3}} gh

2

\sqrt{\frac{4 gh}{3}}

3

\frac{3}{4} gh

4

\sqrt{\frac{3 g h}{4}}

Explanation:

B For pure rolling, total mechanical energy will be conserved.
Friction work $= 0$ (zero)
From figure -

Energy at position $1, (E_{1}) = mgh$
Energy at position 2,
$E_{2} = \frac{1}{2} {mv}_{cm}^{2} + \frac{1}{2} I_{cm} ω^{2}$
$∵ \frac{v_{cm}}{r} = ω and I_{cm} = \frac{{mr}^{2}}{2}$
$E_{2} = \frac{1}{2} m \times r^{2} ω^{2} + \frac{1}{2} \frac{{mr}^{2}}{2} ω^{2}$
$E_{2} = \frac{1}{2} {mr}^{2} ω^{2} + \frac{1}{4} {mr}^{2} ω^{2}$
$E_{2} = \frac{3}{4} {mr}^{2} ω^{2}$
$E_{2} = \frac{3}{4} {mv}_{cm}^{2} (∵ v_{cm} = r ω)$
From law of conservation of energy $E_{1} = E_{2}$
$mgh = \frac{3}{4} {mv}_{cm}^{2}$
$v_{cm}^{2} = \frac{4}{3} gh$
$v_{cm} = \sqrt{\frac{4}{3} gh}$

AP EAMCET-03.09.2021

Laws of Motion

146121 A motor cyclist wants to drive in horizontal circles on the vertical inner surface of a large cylindrical wooden well of radius $8.0 m$ with minimum speed of $5 \sqrt{5} m \cdot s^{- 1}$ . The minimum value of coefficient of friction between the tires and the wall of the well must be - $(g = 10$ $m \cdot s^{- 2}$ )

1 0.10

2 0.64

3 0.30

4 0.40

Explanation:

B Given that-
Radius of wooden well $= 8.0 m$
Minimum speed $= 5 \sqrt{5} m \cdot s^{- 1}$
Friction force, $f_{s} = mg$
Centripetal force, $N = \frac{{mv}^{2}}{r}$
We know,
$μ = \frac{f_{s}}{N}$
$∴ μ = \frac{mg}{\frac{{mv}^{2}}{r}}$
$μ = \frac{r g}{v^{2}}$
$μ = \frac{8 \times 10}{(5 \sqrt{5})^{2}}$
$μ = \frac{80}{125}$
$μ = 0.64$

AP EAMCET-19.08.2021

Laws of Motion

146122 A block B, lying on a table, weighs ' $W$ '. The coefficient of static friction between the block and the table is $μ$ . Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

1

\frac{W \tan θ}{μ}

2

μ W \tan θ

3

μ W \sqrt{1 + \tan^{2} θ}

4

μ W \sin θ

Explanation:

B
The block B is under equilibrium by action of tension force ' $T$ '
Hence $T = μ N$ and $N = W$
So, $T = μ N = μ W$
At knot applying lami's theorem
$\frac{T}{\sin (90^{\circ} + θ)} = \frac{W_{A}}{\sin (180^{\circ} - θ)} = \frac{P}{\sin 90^{\circ}}$
$\frac{T}{\cos θ} = \frac{W_{A}}{\sin θ} = \frac{P}{1}$
$∴ W_{A} = \frac{T \sin θ}{\cos θ} = T \tan θ$
$W_{A} = μ W \tan θ {∵ T = μ W}$

AP EAMCET-23.08.2021

Laws of Motion

146123 A solid flywheel of mass $20 kg$ and radius 120 $mm$ revolves at $600 rpm$ . Find the total force that must applied by the brake so that the flywheel stop in 3 seconds. Given the coefficient of friction between the wheels and brake lining is 0.1 .

1

80 π N

2

24 π N

3

70 π N

4

60 π N

Explanation:

A Given that,
Speed $n = 600 rpm$
Moment of inertia $I = \frac{1}{2} {mR}^{2}$
Radius $R = 120 mm = 0.12 m$
Mass $m = 20 kg$
Coefficient of friction $μ = 0.1$
Angular speed $(ω) = \frac{2 π n}{60} = \frac{2 \times π \times 600}{60} = 20 π red / \sec$
Angular acceleration $(α) = \frac{Δ ω}{Δ t} = \frac{ω}{t} = \frac{20 π}{3} = \frac{20}{3} π red / s^{2}$
Further, torque $τ = I α$
Here, $μ F . R = I α$
$F = \frac{I α}{μ R}$
$F = \frac{\frac{1}{2} m^{2} \times \frac{20 π}{3}}{μ R}$
Or force, $F = \frac{10 π mR}{3 μ}$
Substituting the value, we get-
$F = \frac{10 π \times 20 \times 0.12}{3 \times 0.1}$
$F = 80 π N$

AP EAMCET (Medical)-05.10.2021

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