AP Physics C: Mechanics practice test 5.

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Question 1 |

### A small disk is rotating about an axis perpendicular to the plane of the disc and passing through its center $C$. The disk changed its angular speed from $7 \ rad/s$ to $12 \ rad/s$ in $3 \ s$. What is the angular displacement of the disk in this time period?

$24.3 \ rad$ | |

$30.2 \ rad$ | |

$28.5 \ rad$ | |

$20.7 \ rad$ | |

$19.2 \ rad$ |

Question 1 Explanation:

Angular acceleration $α=\dfrac{12-7}{3}=1.67 \ rad/s^2$

Using $α=1.67 \ rad/s^2, ω_o=7 \ rad/s, \ t=3 s$ in the equation $Δθ=ω_o t+\dfrac{1}{2} αt^2$

$ Δθ=7×3+\dfrac{1}{2}×1.67×3^2=28.5 \ rad$

Using $α=1.67 \ rad/s^2, ω_o=7 \ rad/s, \ t=3 s$ in the equation $Δθ=ω_o t+\dfrac{1}{2} αt^2$

$ Δθ=7×3+\dfrac{1}{2}×1.67×3^2=28.5 \ rad$

Question 2 |

### The below rod is in mechanical equilibrium.

### $C$ represents the center of the rod, and it has a length of $20 \ cm$. Find the magnitude $F$ and the direction $θ$ for the force applied on the right edge.

$F=25 \ N$ and $θ=53.1°$ | |

$F=20 \ N$ and $θ=40.2°$ | |

$F=30 \ N$ and $θ=30.6°$ | |

$F=34 \ N$ and $θ=53.1°$ | |

$F=25 \ N$ and $θ=60.0$° |

Question 2 Explanation:

Resolve the force acting on the right edge as follows:

For translational equilibrium,

$F \cos θ=15$ and $20+F \sin θ=40$

For rotational equilibrium,

$-20×10+F \sinθ×10=0$

$→ F \sin θ=20 \ N$ and $F \cos θ=15 \ N$

$F=\sqrt{20^2+15^2}=25 \ N$

And, $\tan θ=\dfrac{20}{15}=1.33$

Or $θ≈53.1°$

For translational equilibrium,

$F \cos θ=15$ and $20+F \sin θ=40$

For rotational equilibrium,

$-20×10+F \sinθ×10=0$

$→ F \sin θ=20 \ N$ and $F \cos θ=15 \ N$

$F=\sqrt{20^2+15^2}=25 \ N$

And, $\tan θ=\dfrac{20}{15}=1.33$

Or $θ≈53.1°$

Question 3 |

### Two hard spheres of mass $2 \ kg$ each are $1 \ m$ apart. When the distance between the two spheres is increased to $1.5 \ m$, the moment of inertia of the system of two spheres increases to,

### (The axis of rotation passes through the midpoint of the line joining the two spheres)

$2.50 \ kgm^2$ | |

$3.00 \ kgm^2$ | |

$4.00 \ kgm^2$ | |

$3.50 \ kgm^2$ | |

$2.25 \ kgm^2$ |

Question 3 Explanation:

Distance of each sphere from the midpoint of the line joining the two spheres

$=\dfrac{1.5}{2}$

$=0.75 \ m$

Moment of inertia

$=∑mr^2$

$=2×0.75^2+2×0.75^2$

$=2.25 \ kgm^2$

$=\dfrac{1.5}{2}$

$=0.75 \ m$

Moment of inertia

$=∑mr^2$

$=2×0.75^2+2×0.75^2$

$=2.25 \ kgm^2$

Question 4 |

### The angular displacement of a body is given by the formula,

### $θ(t)=2t(1+t^3 )$

### where $θ$ is in $rad$ and $t$ is in $s$. Which graph best represents the angular acceleration of the body?

Question 4 Explanation:

$θ(t)=2t(1+t^3 )=2t+2t^4 $

$ω=\dfrac{d}{dt} θ(t)=2+8t^3 $

$α=\dfrac{d}{dt} ω(t)=24t^2 $

Option D has all these values approximately true

$ω=\dfrac{d}{dt} θ(t)=2+8t^3 $

$α=\dfrac{d}{dt} ω(t)=24t^2 $

$α$ | $0$ | $6$ | $24$ | $54$ |

$t$ | $0$ | $0.5$ | $1$ | $1.5$ |

Question 5 |

### A wheel of radius $4 \ cm$ is rolling on a surface without slipping with an angular velocity of $2 \ rad/s$, as shown in the figure below.

### It is known that the net velocity of point A is the vector sum of the velocity of the center of mass and the (rotational) velocity of point A about the center of mass. What is the magnitude of the net velocity of point $A$?

### (Hint: Velocity of Center of Mass for motion without slipping is $ω × r$)

$13.5 \ cm/s$ | |

$14.8 \ cm/s$ | |

$18.0 \ cm/s$ | |

$12.5 \ cm/s$ | |

$15.7 \ cm/s$ |

Question 5 Explanation:

Velocity of any point on the wheel is equal to the vector sum of the translational and rotational velocity

Velocity of the center of mass $=ω×r=8 \ cm/s$ (along $\skew{2}\hat{i}$)

For point A, Rotational velocity $=ω×r=8 \ cm/s$

And its direction is along the tangent to the circle at point A,

That is along $(\cos45° \skew{2}\hat{i} - \sin 45° \skew{5}\hat{j})=\dfrac{1}{\sqrt{2}} (\skew{2}\hat{i} - \skew{5}\hat{j})$

Net velocity $=\left(8\skew{2}\hat{i} + \dfrac{8}{\sqrt{2}} (\skew{2}\hat{i} - \skew{5}\hat{j} )\right)=(8+4\sqrt{2}) \skew{2}\hat{i} -4\sqrt{2} \skew{5}\hat{j} \ cm/s$

Magnitude of linear velocity $=\sqrt{13.65^2+5.65^2 } = 14.77 \ cm/s $

Velocity of the center of mass $=ω×r=8 \ cm/s$ (along $\skew{2}\hat{i}$)

For point A, Rotational velocity $=ω×r=8 \ cm/s$

And its direction is along the tangent to the circle at point A,

That is along $(\cos45° \skew{2}\hat{i} - \sin 45° \skew{5}\hat{j})=\dfrac{1}{\sqrt{2}} (\skew{2}\hat{i} - \skew{5}\hat{j})$

Net velocity $=\left(8\skew{2}\hat{i} + \dfrac{8}{\sqrt{2}} (\skew{2}\hat{i} - \skew{5}\hat{j} )\right)=(8+4\sqrt{2}) \skew{2}\hat{i} -4\sqrt{2} \skew{5}\hat{j} \ cm/s$

Magnitude of linear velocity $=\sqrt{13.65^2+5.65^2 } = 14.77 \ cm/s $

Question 6 |

### A body of moment of inertia $I$ undergoes rotational motion (starting from rest) due to constant torque $τ$ acting on it. If the angular displacement of the body is equal to $2πN$ (where $N$ is the number of revolutions of the body), find the average time period of the body.

$\sqrt{\dfrac{4I}{Nτ}}$ | |

$\sqrt{\dfrac{2πI}{Nτ}}$ | |

$\sqrt{\dfrac{πI}{Nτ}}$ | |

$\sqrt{\dfrac{8πI}{Nτ}}$ | |

$\sqrt{\dfrac{4πI}{Nτ}}$ |

Question 6 Explanation:

Angular acceleration $α=\dfrac{τ}{I}$

Using $α=\dfrac{τ}{I}, Δθ=2πN$ and $ω_o=0$ in $Δθ=ω_o t+\dfrac{1}{2} \ αt^2$

$2πN=\dfrac{1}{2} \dfrac{τ}{I} \ t^2$

$t=\sqrt{\dfrac{4πNI}{τ}}$

For one revolution,

$\dfrac{t}{N}=\sqrt{\dfrac{4πI}{Nτ}}$

Using $α=\dfrac{τ}{I}, Δθ=2πN$ and $ω_o=0$ in $Δθ=ω_o t+\dfrac{1}{2} \ αt^2$

$2πN=\dfrac{1}{2} \dfrac{τ}{I} \ t^2$

$t=\sqrt{\dfrac{4πNI}{τ}}$

For one revolution,

$\dfrac{t}{N}=\sqrt{\dfrac{4πI}{Nτ}}$

Question 7 |

### Four particles of mass m each are placed at the corners of a rectangle of length $a$ and width $b$. What is the moment of inertia of the system about an axis passing through one of the corners and perpendicular to the plane of the rectangle?

$2m(a^2+b^2 )$ | |

$m(a^2+b^2 )$ | |

$2m(2a^2+b^2 )$ | |

$2m(a^2+2b^2 )$ | |

$3m(a^2+b^2 )$ |

Question 7 Explanation:

The system described in the question is shown below:

The distance of mass m to the left of the axis $=a$

The distance of mass m to the bottom of the axis $=b$

The distance of mass m on the diagonally opposite corner $=\sqrt{a^2+b^2}$

Moment of inertia $I=∑mr^2$

$=m(0)+m(a^2 )+m(b^2 )+m(a^2+b^2 )$

$=2m(a^2+b^2 )$

The distance of mass m to the left of the axis $=a$

The distance of mass m to the bottom of the axis $=b$

The distance of mass m on the diagonally opposite corner $=\sqrt{a^2+b^2}$

Moment of inertia $I=∑mr^2$

$=m(0)+m(a^2 )+m(b^2 )+m(a^2+b^2 )$

$=2m(a^2+b^2 )$

Question 8 |

### Find the angular acceleration of the pulley having a moment of inertia of 0.1 kgm^2 in the figure below.

### (Assume the rope does not slip on the pulley.)

$11.8 \ rad/s^2$ | |

$12.5 \ rad/s^2$ | |

$13.7 \ rad/s^2$ | |

$9.8 \ rad/s^2$ | |

$8.7 \ rad/s^2 $ |

Question 8 Explanation:

From the free body diagrams of the system,

$T_1-3g=3a … (1)$

$7g-T_2=7a … (2)$

$(T_2-T_1 )×0.04=0.1α … (3)$

$a=0.04α … (4)$

Using $(1), (2)$ and $(3)$

$(7g-7a)-(3g+3a)=2.5α$

Using $(4)$ and simplifying gives,

$4g-0.4α=2.5α$

$α=13.7 \ rad/s^2$

$T_1-3g=3a … (1)$

$7g-T_2=7a … (2)$

$(T_2-T_1 )×0.04=0.1α … (3)$

$a=0.04α … (4)$

Using $(1), (2)$ and $(3)$

$(7g-7a)-(3g+3a)=2.5α$

Using $(4)$ and simplifying gives,

$4g-0.4α=2.5α$

$α=13.7 \ rad/s^2$

Question 9 |

### A solid cylinder of mass $2.4 \ kg$ and radius $0.5 \ m$ is rolling on a horizontal surface without slipping. The angular velocity of the cylinder is $4 \ rad/s$. What is the total kinetic energy of the cylinder?

$8.3 \ J$ | |

$7.9 \ J$ | |

$7.2 \ J$ | |

$5.4 \ J$ | |

$6.6 \ J$ |

Question 9 Explanation:

Total $KE=\dfrac{1}{2} MV_{center}^2+\dfrac{1}{2} I_{cylinder} ω^2$

Using $V_{center}=ωR$ and $I_{cylinder}=\dfrac{1}{2} MR^2$ in the above equation gives,

Total $KE=\dfrac{1}{2} Mω^2 R^2+\dfrac{1}{4} Mω^2 R^2=\dfrac{3}{4} Mω^2 R^2$

Using $M=2.4 \ kg, ω=4 \ rad/s$ and $R=0.5 \ m$ gives,

Total $KE=\dfrac{3}{4}×2.4×4^2×(0.5)^2=7.2 \ J$

Using $V_{center}=ωR$ and $I_{cylinder}=\dfrac{1}{2} MR^2$ in the above equation gives,

Total $KE=\dfrac{1}{2} Mω^2 R^2+\dfrac{1}{4} Mω^2 R^2=\dfrac{3}{4} Mω^2 R^2$

Using $M=2.4 \ kg, ω=4 \ rad/s$ and $R=0.5 \ m$ gives,

Total $KE=\dfrac{3}{4}×2.4×4^2×(0.5)^2=7.2 \ J$

Question 10 |

### A non-uniform rod of mass density $ρ(x)=2Ax \ g/cm$ can rotate about point $Y$, as shown below.

### What is the moment of inertia of the rod about point $Y$?

### (Here, $x$ is the distance from the origin $O$, and $A$ is a positive constant)

$\dfrac{Al^3}{2}$ | |

$\dfrac{Al^4}{6}$ | |

$\dfrac{5Al^4}{6}$ | |

$\dfrac{Al^2}{12}$ | |

$\dfrac{Al^4}{4}$ |

Question 10 Explanation:

$I=\int r^2 \ dm$

Using $dm=ρ(x)dx $ and $ r=(l-x)$ gives,

$I=\int_0^l ρ(x) (l-x)^2 \ dx$

$=\int_0^l 2Ax(l-x)^2 \ dx$

$=2A\int_0^l x(l^2-2xl+x^2 ) \ dx$

$=2A\int_0^l (l^2 x-2x^2 l+x^3 ) \ dx$

$=2A \left( \dfrac{l^2 x^2}{2}-\dfrac{2}{3} x^3 l+ \ \dfrac{x^4}{4} \right) |_0^l$

$=2A \left( \dfrac{l^4}{2} - \dfrac{2l^4}{3}+\dfrac{l^4}{4}\right)$

$=\dfrac{Al^4}{6}$

Using $dm=ρ(x)dx $ and $ r=(l-x)$ gives,

$I=\int_0^l ρ(x) (l-x)^2 \ dx$

$=\int_0^l 2Ax(l-x)^2 \ dx$

$=2A\int_0^l x(l^2-2xl+x^2 ) \ dx$

$=2A\int_0^l (l^2 x-2x^2 l+x^3 ) \ dx$

$=2A \left( \dfrac{l^2 x^2}{2}-\dfrac{2}{3} x^3 l+ \ \dfrac{x^4}{4} \right) |_0^l$

$=2A \left( \dfrac{l^4}{2} - \dfrac{2l^4}{3}+\dfrac{l^4}{4}\right)$

$=\dfrac{Al^4}{6}$

Question 11 |

### The moment of inertia of a solid sphere of mass $M$ and radius $R$ about an axis passing through its center is equal to $ \dfrac{2}{5} \ MR^2$. What would be the moment of inertia for an axis tangent to the sphere?

$\dfrac{2}{5} \ MR^2$ | |

$\dfrac{3}{5} \ MR^2$ | |

$\dfrac{4}{5} \ MR^2$ | |

$\dfrac{7}{5} \ MR^2$ | |

$\dfrac{8}{5} \ MR^2$ |

Question 11 Explanation:

From parallel axis theorem,

$I_{tan}=I_{cm}+Md^2$

Using $I_{cm}=\dfrac{2}{5} \ MR^2$ and $d=R$

$I_{tan}=\dfrac{2}{5} \ MR^2+MR^2=\dfrac{7}{5} \ MR^2$

$I_{tan}=I_{cm}+Md^2$

Using $I_{cm}=\dfrac{2}{5} \ MR^2$ and $d=R$

$I_{tan}=\dfrac{2}{5} \ MR^2+MR^2=\dfrac{7}{5} \ MR^2$

Question 12 |

### The diagram below shows the directions in which two particles of mass $m$ each move in an $x-y$ plane.

### The magnitude of the velocities of the two particles, $X$ and $Y$, are $5 \ m/s$ and $6 \ m/s$, respectively. Determine the angular momentum of $X$ with respect to $Y$. (Hint: What is the velocity of $X$ with respect to $Y$?)

$(-34.96m) k ̂ \ m/s$ | |

$(-69.62m) k ̂ \ m/s$ | |

$(-39.42m) k ̂ \ m/s$ | |

$(-36.63m) k ̂ \ m/s$ | |

$(-63.16m) k ̂ \ m/s$ |

Question 12 Explanation:

Velocity of $X=-5\skew{5}\hat{j} \ m/s$

Velocity of $Y=6(\cos 45° \skew{2}\hat{i} + \sin 45° \skew{5}\hat{j} )=3\sqrt{2} (\skew{2}\hat{i}+j ̂ ) \ m/s$

Velocity of $X$ with respect to $Y=-5\skew{5}\hat{j}-3\sqrt{2} (\skew{2}\hat{i}+\skew{5}\hat{j})=-(4.24\skew{2}\hat{i}+9.24\skew{5}\hat{j} ) \ m/s$

Position of $X$ with respect to $Y =(7\skew{2}\hat{i}+3\skew{5}\hat{j} )-(2\skew{2}\hat{i}+7\skew{5}\hat{j} )=(5\skew{2}\hat{i}-4\skew{5}\hat{j} ) \ m/s$

Angular momentum $\overrightarrow{L}=m(\overrightarrow{r}×\overrightarrow{v})$

$=-m(5×9.24+4×4.24) k ̂ $

$=m(-63.16) k ̂ \ m/s$

Velocity of $Y=6(\cos 45° \skew{2}\hat{i} + \sin 45° \skew{5}\hat{j} )=3\sqrt{2} (\skew{2}\hat{i}+j ̂ ) \ m/s$

Velocity of $X$ with respect to $Y=-5\skew{5}\hat{j}-3\sqrt{2} (\skew{2}\hat{i}+\skew{5}\hat{j})=-(4.24\skew{2}\hat{i}+9.24\skew{5}\hat{j} ) \ m/s$

Position of $X$ with respect to $Y =(7\skew{2}\hat{i}+3\skew{5}\hat{j} )-(2\skew{2}\hat{i}+7\skew{5}\hat{j} )=(5\skew{2}\hat{i}-4\skew{5}\hat{j} ) \ m/s$

Angular momentum $\overrightarrow{L}=m(\overrightarrow{r}×\overrightarrow{v})$

$=-m(5×9.24+4×4.24) k ̂ $

$=m(-63.16) k ̂ \ m/s$

Question 13 |

### Questions 13 and 14 are based on the below information:

### A torque $τ(t) = 6t^2$ acts on a thin disc of moment of inertia $3 \ kgm^2$. The disc moves about an axis, passing through its center of mass and perpendicular to the plane of the disc.

### What is the change in the angular momentum of the disc when it moves from time $t=2 \ s$ to $t=5 \ s$?

$207 \ kgm^2/s$ | |

$216 \ kgm^2/s$ | |

$234 \ kgm^2/s$ | |

$125 \ kgm^2/s$ | |

$192 \ kgm^2/s$ |

Question 13 Explanation:

$ΔL=\intτ(t)dt$

Thus,

$ΔL=\int_2^5 6t^2 \ dt$

$=6\int_2^5 t^2 \ dt$

$=6 \dfrac{t^3}{3} |_2^5$

$=2 (5^3-2^3 )$

$=234 \ kgm^2/s $

Thus,

$ΔL=\int_2^5 6t^2 \ dt$

$=6\int_2^5 t^2 \ dt$

$=6 \dfrac{t^3}{3} |_2^5$

$=2 (5^3-2^3 )$

$=234 \ kgm^2/s $

Question 14 |

### A torque $τ(t) = 6t^2$ acts on a thin disc of moment of inertia $3 \ kgm^2$. The disc moves about an axis, passing through its center of mass and perpendicular to the plane of the disc.

### If the disc was momentarily at rest at time $t=2 \ s$, what is the angular displacement of the disc in the time interval $t=2 \ s$ to $t=5 \ s$?

$92.5 \ rad$ | |

$85.5 \ rad$ | |

$99.8 \ rad$ | |

$82.7 \ rad$ | |

$88.2 \ rad$ |

Question 14 Explanation:

$α(t)=\dfrac{τ(t)}{I}= 2t ^2$

$ω(t)=\int 2t^2 \ dt$

$=\dfrac{2}{3} t^3+C$

Using $ω(2)=0 \ rad/s$ gives $C=-\dfrac{16}{3} $

$θ(t)=\int \dfrac{2}{3} t^3-\dfrac{16}{3} \ dt $

$=\dfrac{1}{6} t^4- \dfrac{16}{3} t+C'$

$Δθ=θ(5)-θ(2)=\dfrac{1}{6} (5^4-2^4 )-\dfrac{16}{3}(5-2)$

$=85.5 \ rad $

$ω(t)=\int 2t^2 \ dt$

$=\dfrac{2}{3} t^3+C$

Using $ω(2)=0 \ rad/s$ gives $C=-\dfrac{16}{3} $

$θ(t)=\int \dfrac{2}{3} t^3-\dfrac{16}{3} \ dt $

$=\dfrac{1}{6} t^4- \dfrac{16}{3} t+C'$

$Δθ=θ(5)-θ(2)=\dfrac{1}{6} (5^4-2^4 )-\dfrac{16}{3}(5-2)$

$=85.5 \ rad $

Question 15 |

### Two solid discs are rotating about a common axis of rotation at speeds $8.5 \ rad/s$ and $-12.5 \ rad/s$. The masses of the discs are $4 \ kg$ and $3 \ kg$, respectively. The radius of both the discs is the same and is equal to $20 \ cm$.

### If the two discs are combined to rotate together as shown in the diagram above, then what is the common angular velocity of the two discs?

### (Hint: No external torque is needed to achieve this)

$-0.50 \ rad/s$ | |

$3.20 \ rad/s$ | |

$0.25 \ rad/s$ | |

$1.75 \ rad/s$ | |

$-1.35 \ rad/s$ |

Question 15 Explanation:

Before the discs are combined,

Net Angular Momentum $=I_1 ω_1+I_2 ω_2$

$=\dfrac{1}{2} \ m_1 r^2 ω_1+\dfrac{1}{2} \ m_2 r^2 ω_2$

$=\dfrac{1}{2} \ r^2 (m_1 ω_1+m_2 ω_2 )$

$=\dfrac{1}{2} \ (0.2^2 )(4×8.5-3×12.5)$

$=-0.07 \ kgm^2/s$

After the discs are combined,

Net Angular Momentum $=Iω$

$=\dfrac{1}{2} \ mr^2 \ ω $

$=\dfrac{1}{2} (7)(0.2^2)ω$

$=0.14ω$

$=-0.07$

Thus, $ω=-0.50 \ rad/s$ ($-$ sign shows the velocity is anticlockwise)

Net Angular Momentum $=I_1 ω_1+I_2 ω_2$

$=\dfrac{1}{2} \ m_1 r^2 ω_1+\dfrac{1}{2} \ m_2 r^2 ω_2$

$=\dfrac{1}{2} \ r^2 (m_1 ω_1+m_2 ω_2 )$

$=\dfrac{1}{2} \ (0.2^2 )(4×8.5-3×12.5)$

$=-0.07 \ kgm^2/s$

After the discs are combined,

Net Angular Momentum $=Iω$

$=\dfrac{1}{2} \ mr^2 \ ω $

$=\dfrac{1}{2} (7)(0.2^2)ω$

$=0.14ω$

$=-0.07$

Thus, $ω=-0.50 \ rad/s$ ($-$ sign shows the velocity is anticlockwise)

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