Question

A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can rotate with negligible friction about a stationary horizontal axis. The drum is not a uniform cylinder and has unknown moment of inertia. When you release the bucket from rest, you find that it has a downward acceleration of magnitude (a). What is the tension in the cable between the drum and the bucket? (b) What is the moment of inertia of the drum about its rotation axis?

c) In the primitive yo-yo apparatus (Figure 1), you replace the solid cylinder with a hollow cylinder of mass M, outer radius R, and inner radius R/2. Find the magnitude of the downward acceleration of the hollow cylinder. what is the Tension?

Answer 1

Answer:

Explanation:

Let T be the tension .

Applying newton's second law on the downward movement of the bucket

mg - T = ma

On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum

TR = Iα

TR = Ia/ R

T =  Ia/ R²

Replacing this value of T in the other equation

mg - T = ma

mg - Ia/ R²  = ma

mg =  Ia/ R² +ma

a ( I/ R² +m)= mg

a = mg / ( I/ R² +m)

mg - T = ma

mg - ma  = T

mg - m x mg / ( I/ R² +m) = T

mg - m²g / ( I/ R² +m ) = T

mg - mg / ( 1 + I / m R² ) = T

b ) T =  Ia/ R²

I = TR² / a

c ) Moment of inertia of hollow cylinder

I = 1/2  M ( R² - R² / 4 )

= 3/4 x 1/2 MR²

= 3/8 MR²

I / R² = 3/8 M

a = mg / ( I/ R² +m)

a = mg / ( 3/8 M + m )

T =  Ia/ R²

= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²

= $\frac{3mMg}{(3M +8m)}$

Answer 2

• The tension in the cable between the drum and the bucket-

$T = mg - (m²g/(I/R² + m)$

Let T be the tension in the cable between the drum and the bucket.

What is the moment of inertia?

The quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of torque (turning force).

Applying newton's second law on the downward movement of the bucket.

$mg - T = ma$

On the drum, a torque of TR will be acting which will create an angular acceleration of α in it. If I a moment of inertia of the drum.

$TR = Iα$

$TR = Ia/ R$

$T = Ia/ R²$

Replacing this value of T in the other equation.

$mg - T = ma$

$mg - Ia/ R² = ma$

$mg = Ia/ R² +ma$

$a ( I/ R² +m)= mg$

$a = mg / ( I/ R² +m)$

$mg - T = ma$

$mg - ma = T$

$mg - m x mg / ( I/ R² +m) = T$

$mg - m²g / ( I/ R² +m ) = T$

$mg - mg / ( 1 + I / m R² ) = T$

b ) $T = Ia/ R²$

$I = TR² / a$

• Moment of inertia of the hollow cylinder.

$I = 1/2 M ( R² - R² / 4 )$

$= 3/4 x 1/2 MR²$

$= 3/8 MR²$

$I / R² = 3/8 M$

$a = mg / ( I/ R² +m)$

$a = mg / ( 3/8 M + m )$

$T = Ia/ R²$

=$3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²$

= ($\frac{ 3mMg}{3M+8m}$)

Therefore, the result is = ($\frac{ 3mMg}{3M+8m}$).

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