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e-lub [12.9K]
3 years ago
5

During a hard drive crash the read/write head scrapes against the disk with a coefficient of kinetic friction of µk and normal f

orce of N. Assume that just before the head crash the disk is rotating at ω0 rad/s, and the distance of the head from the disk axis is r. You can ignore any friction at the bearing (rotational axis of the disk). Assume that the disk is uniform, and has radius R and mass M. What is the angular acceleration associated with the torque from the crashed disk head?
Physics
1 answer:
Alex_Xolod [135]3 years ago
3 0

Answer:

 α = \frac{2 \mu \  N}{m \ r}

Explanation:

For this exercise we use Newton's equation for rotational motion

           ∑ τ = I α

the troque is

           α = Fr .r

the moment of inertia of a cylinder is

           I = ½ m r²

we substitute

         fr r = (½ m r²) α

the expression friction is

         fr = μ N

we substitute

         μ N r = ½ m r² α

       

         α = \frac{2 \mu \  N}{m \ r}

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The surface can only withstand a pressure Face A (50cm × 30cm) and Face C (50cm × 15cm) as the surface can only withstand a pressure of 0.25 N/cm³.

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A box has the dimensions of 50 cm × 30 cm × 15 cm

Let each face be A, B and C

The weight of the box = 150 N

Formula for pressure is

P = F/A

To find out which face of the box can withstand a pressure of 0.25 N/cm

we need find the area of each face and find its pleasure

Face A = 50 cm × 30 cm

Area A = l × b

            = 50 × 30

            = 1500 cm²

Pressure A = 150/1500

                  = 0.1 N/cm³

                    0.25 > 0.1

The surface can definitively withstand the pressure of Face A

Face B = 30 cm × 15 cm

Area A = l × b

            = 30 × 15

            = 450 cm²

Pressure A = 150/450

                  = 0.3 N/cm³

                    0.25 < 0.3

The surface could not withstand the pressure of Face B

Face C = 50 cm × 15 cm

Area A = l × b

            = 50 × 15

            = 750 cm²

Pressure A = 150/750

                  = 0.2 N/cm³

                    0.25 > 0.2

The surface can definitively withstand the pressure of Face C

Thus, The surface can only withstand a pressure Face A (50cm × 30cm) and Face C (50cm × 15cm) as the surface can only withstand a pressure of 0.25 N/cm³.

Learn more about Pressure

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