Question

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 kg, are welded together and mounted on a frictionless axis through their common center. A light string is wrapped around the edge of the smaller disk and a 1.60 kg block is suspended from the free end of the string. (a) What is the magnitude of the downward acceleration of the block after it is released? (b) Repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk.

Answer 1

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

• Mass of the larger disk = $M_2\ =\ 1.60\ kg$
• Mass of the smaller disk = $M_1\ =\ 0.900\ kg$
• Radius of the larger disk = $R_2\ =\ 5.00\ cm\ =\ 0.05\ m$
• Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$
• Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

$\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

$mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)$

From the pulley,

$\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)$

From the equation (1) and (2),

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2$

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = $R_2\ =\ 0.05\ m$

Let $a_2$ be the acceleration of the block while connecting to the larger pulley.$\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2$

Answer 2

The magnitude of the downward acceleration of the block after it is released is 2.912 m/s². The downward acceleration of the system when the string is wrapped around the edge of the larger disk is 6.25 m/s².

From the given information;

The moment of inertia for the combined two metal disks system can be computed as:

$\mathbf{I_{system}=\dfrac{1}{2} M_1R_1^2 + \dfrac{1}{2}M_2R_2^2}$

where;

• M1 = 0.900 kg
• R1 = 2.45 cm = 0.0245 m
• M2 = 1.60 kg
• R2 = 5.00 cm = 0.05 m
• mass of block (m) = 1.60 kg

∴

$\mathbf{I_{system}=\dfrac{1}{2} (0.9 \ kg) (0.0245\ m )^2 + \dfrac{1}{2}(1.60 ) (0.05\ m) ^2}$

$\mathbf{I_{system}=0.0022701 \ kg.m^2}$

Now, using the equation for the torque acting on the disk;

$\mathbf{Torque \ (\tau) = I_{system }\times \alpha}$

where;

• angular acceleration ∝ $\mathbf{= \dfrac{a}{R_1}}$

As such, replacing it into the previous equation for the torque, we get:

$\mathbf{\tau = I_{system } \dfrac{a}{R_1}}$

For the smaller disk, the torque can be computed as;

$\mathbf{\tau = TR_1}$

here;

• T = Tension in the string

∴

equating both torque together, we have:

$\mathbf{I_{system } \dfrac{a}{R_1}= TR_1}$

$\mathbf{I_{system }\times a= TR_1^2}$

Making the tension T the subject, we have;

$\mathbf{T = \dfrac{I_{system} \times a}{R_1^2} }$

However, if we look at the block from the given information, the net force acting on the block can be expressed as:

• mg - T = ma

where;

• T = tension

∴

$\mathbf{mg - \dfrac{I_{system }\times a }{R_1^2} = ma }$

$\mathbf{mg = ma+ \dfrac{I_{system }\times a }{R_1^2} }$

$\mathbf{mg = a\Big (m + \dfrac{I_{system } }{R_1^2} \Big) }$

Making acceleration (a) the subject, we have:

$\mathbf{a = \Big ( \dfrac{mg}{m +\dfrac{I_{system}}{R_1^2} } \Big )}$

$\mathbf{a = \Big ( \dfrac{1.60 \ kg \times 9.8 \ m/s^2 }{1.60 \ kg +\dfrac{0.0022701}{( 0.0245)^2} } \Big )}$

a = 2.912 m/s²

If the string is wrapped around the edge to the larger disk (R2), then, the downward acceleration of the system is calculated by using the formula:

$\mathbf{a = \Big ( \dfrac{mg}{m +\dfrac{I_{system}}{R_2^2} } \Big )}$

$\mathbf{a = \Big ( \dfrac{1.60 \ kg \times 9.8 \ m/s^2 }{1.60 \ kg +\dfrac{0.0022701}{( 0.05)^2} } \Big )}$

a = 6.25 m/s²

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