I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.

The relation between the linear acceleration and the angular acceleration is

a = R α .... (1)

According to the diagram,

T x R = I x α

T x R = I x a / R from equation (1)

T = I x a / R² .... (2)

mg - T = ma .... (3)

Substitute the value of T from equation (2) in equation (3)

$mg - \frac{Ia}{R^{2}}=ma$

$a=\frac{mg}{m+\frac{I}{R^{2}}}$

T is the acceleration in the system

Substitute the value of a in equation (2)

$T = \frac{I}{R^{2}}\times \frac{mg}{m+\frac{I}{R^{2}}}$

$T=\frac{I\times mg}{I+mR^{2}}$

This is the tension in the string.

4 0

3 years ago

A DNA macromolecule is made up of two strands of DNA, which are connected to form a DNA double helix. The bonds between

Arturiano [62]

Answer:

where is the multiple choice?

6 0

3 years ago

Air passing over an airplane's wing travels ____________________, and therefore exerts ____________________ pressure than air tr

Aleksandr [31]

<span>c. faster, less

is the answer
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6 0

3 years ago

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Suppose I have a vector that is 7 units long and that makes an angle of +30 degrees from the positive x-axis. I want to add to t

Vinil7 [7]

Answer:

sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

Explanation:

We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis

So x component of the vector $=7cos30^{\circ}=7\times 0.866=6.06$

y component of the vector $=7sin30^{\circ}=7\times 0.5=3.5$

So vector will be 6.06i+3.5j

Now other vector of length of 7 units and makes an angle of 120° with positive x-axis

So x component of vector $=7cos120^{\circ}=7\times -0.5=-3.5i$

y component of the vector $=7sin120^{\circ}=7\times 0.866=6.06j$

Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

5 0

3 years ago