Question

Expressions have been found for the vertical acceleration of the cylinder ay and the angular acceleration α of the cylinder in the k^ direction; both expressions include an unknown variable, namely, the tension T in the vertical section of string. The string constrains the rotational and vertical motions, providing a third equation relating ay and α. Solve these three equations to find the vertical acceleration, ay, of the center of mass of the cylinder. Express ay in terms of g, m, r, and I; a positive answer indicates upward acceleration.

Answer 1

Complete Question

A cylinder with a moment of inertia I (about its axis of symmetry), mass m, and radius r has a mass-less string wrapped around it which is tied to the ceiling .

At time t=0 the cylinder is released from rest at height h above the ground. Use g for the magnitude of the acceleration of gravity. Assume that the string does not slip on the cylinder. Let v represent the instantaneous velocity of the center of mass of the cylinder, and let
`\omega` represent the instantaneous angular velocity of the cylinder about its center of mass. Note that there are no horizontal forces present, so for this problem
`\= v =- v\r j \ and \ \omega =-\omega \r k.`

Expressions have been found for the vertical acceleration of the cylinder
`a_y` and the angular acceleration α of the cylinder in the k^ direction; both expressions include an unknown variable, namely, the tension T in the vertical section of string. The string constrains the rotational and vertical motions, providing a third equation relating
`a_y` and α. Solve these three equations to find the vertical acceleration,
`a_y`, of the center of mass of the cylinder. Express
`a_y` in terms of g, m, r, and I; a positive answer indicates upward acceleration.

Answer:

The vertical acceleration is
`a_y = (mg)/([m+ (I)/(r^2) ])`

Step-by-step explanation:

The equation of motion is mathematically represented as

`ma_y = T -mg ---(1)`

The relation between the tension and the moment of inertia is

`I \alpha = -Tr`

`T = (-I\alpha )/(r) ---(2)`

Now angular acceleration can be mathematically represented as

`\alpha = (a_y)/(r)`

Now substituting this into equation 2

`T = (-Ia_y)/(r^2) ---(3)`

Now substituting these into equation 1

`ma_y = (-Ia_y)/(r^2) - mg`

`a_y [m + (I)/(r^2) ] = -mg`

Hence the vertical acceleration is evaluates as

`a_y = (mg)/([m+ (I)/(r^2) ])`

Answer 2

The first part of this question is missing and it says;

A cylinder with a moment of inertia I (about its axis of symmetry), mass m, and radius r has a massless string wrapped around it which is tied to the ceiling(Figure 1) .

At time t=0 the cylinder is released from rest at height h above the ground. Use g for the magnitude of the acceleration of gravity. Assume that the string does not slip on the cylinder. Let v⃗ represent the instantaneous velocity of the center of mass of the cylinder, and let ω⃗ represent the instantaneous angular velocity of the cylinder about its center of mass. Note that there are no horizontal forces present, so for this problem v⃗ =−vj^ and ω⃗ =−ωk^.

Answer:

ay = -mg / (m + I/r²)

Step-by-step explanation:

Therefore we need to find an expression for T in terms of g, m, r, and I.

Equation of motion says;

may = T - mg

The relationship between tension and moment of inertia is

Iα = -Tr

Tr = -Iα

T = -Iα/r

Since

Angular acceleration is given as;

α = ay/r

Thus;

Since, T = T = -Iα/r

We now have,

T = -I(ay/r)/r = -(Iay)/r²

Now, since may = T - mg

We now have;

may = -(Iay)/r² - mg

may + (Iay)/r² = -mg

ay(m + I/r²) = -mg

ay = -mg / (m + I/r²)