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A cable passes over a pulley. Because the cable grips the pulley and the pulley has nonzero mass, the tension in the cable is not the same on opposite sides of the pulley. The force on one side is 137 N, and the force on the other side is 43 N. Assuming that the pulley is a uniform disk of mass 1.21 kg and radius 0.723 m, find the magnitude of its angular acceleration. [For a uniform disk, I = (1/2)mr2 .] Answer in units of rad/s 2

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lublana

Answer:

[tex]\alpha=214.8 rad/s^2[/tex]

Explanation:

We are given that

[tex]F_1=137 N[/tex]

[tex]F_2=43 N[/tex]

Net force=F=[tex]F_1-F_2=137-43=94 N[/tex]

Mass,m=1.21 kg

Radius,r=0.723 m

We have to find the magnitude of its angular acceleration.

Moment of inertia ,[tex]I=\frac{1}{2}mr^2[/tex]

Substitute the values

Torque ,[tex]\tau=I\alpha[/tex]

[tex]F_{net}\times r=\frac{1}{2}mr^2\alpha[/tex]

[tex]\alpha=\frac{2F_{net}}{mr}[/tex]

[tex]\alpha=\frac{2\times 94}{1.21\times 0.723}[/tex]

[tex]\alpha=214.8 rad/s^2[/tex]

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