Question

I hate turntable questions I can not understand this one atallConsider a turntable to be a circular disk of moment ofinertia $I_t$ rotating at a constant angular velocity $omega_i$ around an axis through the center and perpendicular tothe plane of the disk (the disk's "primary axis of symmetry"). Theaxis of the disk is verticaland the disk is supported byfrictionless bearings. The motor of the turntable is off, so thereis no external torque being applied to the axis.
Another disk (a record) is dropped onto the first such that itlands coaxially (the axes coincide). The moment of inertia of therecord is $I_r$ . The initial angular velocity of the second disk iszero.

There is friction between the two disks.
What is the final angular velocity, $omega_f$ , of thetwo disks?Express $omega_f$ in terms of $I_t$ , $I_r$ , and $omega_i$ .

After this "rotational collision," the disks will eventually rotatewith the same angular velocity.Because of friction, rotational kineticenergy is not conserved while the disks' surfaces slip over eachother. What is the final rotational kinetic energy, $K_f$ , of the two spinning disks?Express the final kinetic energy interms of $I_t$ , $I_r$ , and the initial kinetic energy $K_i$ of the two-disk system. No angular velocitiesshouldappear in your answer.

Answer 1

Given thatthe moment of inertia of the first disk is $I_t$ Initialangular velocity of the first disk is ω_ithe moment ofinertia of the2nd disk is $I_r$ .Initialangular velocity of the2nd disk is0---------------------------------------------------------------------------From the conservation ofangularmomentumIt*ωi + Ir*0 = ( It +Ir)ω_fIt*ωi= ( It +Ir)ω_fωf= It*ωi/ ( It + Ir)The final kinetic energy is Kf =(1/2)( It + Ir)ω_f^{2Theinitial kineticenergy is Ki= (1/2)(It)ω_i2}

Answer 2

General guidance

Concepts and reason

The concepts required to solve this problem are the angular momentum, moment of inertia, torque, and rotational kinematics.

Initially, calculate the final angular velocity of the two disks by using the conservation of angular momentum. Next, calculate the final rotational kinetic energy of the two spinning disks by using the angular frequency. Finally, calculate the average torque acting on the bottom disk due to the friction by using the relation between angular acceleration and angular frequency.

Fundamentals

The expression of the angular momentum is,

$L = I\\omega$

Here, L is the angular momentum, I is the moment of inertia, and $\\omega$ is the angular velocity.

The expression of the rotational kinetic energy K is,

$K = \\frac{1}{2}I{\\omega ^2}$

Here, I is the moment of inertia and $\\omega$ is the angular velocity.

The expression of the torque is,

$\\tau = I\\alpha$

Here, $\\tau$ is the torque and $\\alpha$ is the angular acceleration.

The rotational kinematic equation for the angular speed is,

${\\omega _{\\rm{f}}} = {\\omega _{\\rm{i}}} + \\alpha \\Delta t$

Here, ${\\omega _{\\rm{f}}}$ is the final angular speed, ${\\omega _{\\rm{i}}}$ is the initial angular speed, $\\alpha$ is the angular acceleration, and $\\Delta t$ is the time.

Step-by-step

Step 1 of 3

(A)

The expression of the initial angular momentum is,

${L_{\\rm{i}}} = {I_{\\rm{t}}}{\\omega _{\\rm{i}}} + {I_{\\rm{r}}}{\\omega _{\\rm{i}}}'$

Here, ${I_{\\rm{t}}}$ is the moment of inertia of the first disk and ${I_{\\rm{r}}}$ is the moment of inertia of the second disk.

Substitute 0 for ${\\omega _{\\rm{i}}}'$.

$\\begin{array}{c}\\\\{L_{\\rm{i}}} = {I_{\\rm{t}}}{\\omega _{\\rm{i}}} + {I_{\\rm{r}}}\\left( 0 \\right)\\\\\\\\{L_{\\rm{i}}} = {I_{\\rm{t}}}{\\omega _{\\rm{i}}}\\\\\\end{array}$

The expression of the final angular momentum ${L_{\\rm{f}}}$ is,

$\\begin{array}{c}\\\\{L_{\\rm{f}}} = {I_{\\rm{t}}}{\\omega _{\\rm{f}}} + {I_{\\rm{r}}}{\\omega _{\\rm{f}}}\\\\\\\\{L_{\\rm{f}}} = \\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right){\\omega _{\\rm{f}}}\\\\\\end{array}$

Here, ${I_{\\rm{t}}}$ is the moment of inertia of the first disk and ${I_{\\rm{\\tau }}}$ is the moment of inertia of the second disk.

Apply conservation of angular momentum.

${L_{\\rm{i}}} = {L_{\\rm{f}}}$

Substitute ${I_{\\rm{t}}}{\\omega _{\\rm{i}}}$ for ${L_{\\rm{i}}}$ and $\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right){\\omega _{\\rm{f}}}$ for ${L_{\\rm{f}}}$.

${I_{\\rm{t}}}{\\omega _{\\rm{i}}} = \\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right){\\omega _{\\rm{f}}}$

Rearrange for ${\\omega _{\\rm{f}}}$.

${\\omega _{\\rm{f}}} = \\frac{{{I_{\\rm{t}}}{\\omega _{\\rm{i}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}$

Part A

The final angular velocity of the two disks is $\\frac{{{I_{\\rm{t}}}{\\omega _{\\rm{i}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}$.

Explanation | Common mistakes | Hint for next step

The conservation of the angular momentum states that the initial angular momentum of a system is equal to the final angular momentum of the system. The initial angular speed of the second disk is zero that why the initial angular momentum is equal to ${I_{\\rm{t}}}{\\omega _{\\rm{i}}}$.

Step 2 of 3

(B)

The expression of the initial rotational kinetic energy ${K_{\\rm{i}}}$ is,

${K_{\\rm{i}}} = \\frac{1}{2}{I_{\\rm{t}}}{\\omega _{\\rm{i}}}^2$

The expression of the final rotational kinetic energy ${K_{\\rm{f}}}$ is,

${K_{\\rm{f}}} = \\frac{1}{2}\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right){\\omega _{\\rm{f}}}^2$

Substitute $\\frac{{{I_{\\rm{t}}}{\\omega _{\\rm{i}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}$ for ${\\omega _{\\rm{f}}}$.

$\\begin{array}{c}\\\\{K_{\\rm{f}}} = \\frac{1}{2}\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right){\\left( {\\frac{{{I_{\\rm{t}}}{\\omega _{\\rm{i}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}} \\right)^2}\\\\\\\\ = \\frac{1}{2}\\frac{{{I_{\\rm{t}}}^2{\\omega _{\\rm{i}}}^2}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}\\\\\\\\ = \\frac{{{I_{\\rm{t}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}\\left( {\\frac{1}{2}{I_{\\rm{t}}}{\\omega _{\\rm{i}}}^2} \\right)\\\\\\end{array}$

Substitute ${K_{\\rm{i}}}$ for $\\frac{1}{2}{I_{\\rm{t}}}{\\omega _{\\rm{i}}}^2$.

${K_{\\rm{f}}} = \\frac{{{I_{\\rm{t}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}{K_{\\rm{i}}}$

Part B

The final rotational kinetic energy of the two spinning disks is $\\frac{{{I_{\\rm{t}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}{K_{\\rm{i}}}$.

Explanation | Common mistakes | Hint for next step

The initial angular speed of the second disk is zero that why the initial rotational kinetic energy is equal to momentum is equal to $\\frac{1}{2}{I_{\\rm{t}}}{\\omega _{\\rm{i}}}^2$. The final rotational kinetic energy can be expressed in terms of the moment of inertia and the initial kinetic energy and it can be shown by $\\frac{{{I_{\\rm{t}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}{K_{\\rm{i}}}$.

Step 3 of 3

(C)

The expression of the torque is,

$\\tau = {I_{\\rm{t}}}\\alpha$

The rotational kinematic equation for the angular speed is,

${\\omega _{\\rm{f}}} = {\\omega _{\\rm{i}}} + \\alpha \\Delta t$

Rearrange the equation for $\\alpha$.

$\\alpha = \\frac{{{\\omega _{\\rm{f}}} - {\\omega _{\\rm{i}}}}}{{\\Delta t}}$

Substitute $\\frac{{{\\omega _{\\rm{f}}} - {\\omega _{\\rm{i}}}}}{{\\Delta t}}$ for $\\alpha$ in the expression of the torque $\\tau$.

$\\tau = {I_{\\rm{t}}}\\left( {\\frac{{{\\omega _{\\rm{f}}} - {\\omega _{\\rm{i}}}}}{{\\Delta t}}} \\right)$

Part C

The average torque acting on the bottom disk due to friction torque is ${I_{\\rm{t}}}\\left( {\\frac{{{\\omega _{\\rm{f}}} - {\\omega _{\\rm{i}}}}}{{\\Delta t}}} \\right)$.

Explanation

The torque is defined as the product of the moment of inertia and the angular acceleration. Angular acceleration is the change in the angular velocity that a spinning object undergoes per unit time.

Answer

Part A

The final angular velocity of the two disks is $\\frac{{{I_{\\rm{t}}}{\\omega _{\\rm{i}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}$.

Part B

The final rotational kinetic energy of the two spinning disks is $\\frac{{{I_{\\rm{t}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}{K_{\\rm{i}}}$.

Part C

The average torque acting on the bottom disk due to friction torque is ${I_{\\rm{t}}}\\left( {\\frac{{{\\omega _{\\rm{f}}} - {\\omega _{\\rm{i}}}}}{{\\Delta t}}} \\right)$.

Answer only

Part A

The final angular velocity of the two disks is $\\frac{{{I_{\\rm{t}}}{\\omega _{\\rm{i}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}$.

Part B

The final rotational kinetic energy of the two spinning disks is $\\frac{{{I_{\\rm{t}}}}}{{\\left( {{I_{\\rm{t}}} + {I_{\\rm{r}}}} \\right)}}{K_{\\rm{i}}}$.

Part C

The average torque acting on the bottom disk due to friction torque is ${I_{\\rm{t}}}\\left( {\\frac{{{\\omega _{\\rm{f}}} - {\\omega _{\\rm{i}}}}}{{\\Delta t}}} \\right)$.