A mass of 0.34 kg is fixed to the end of a 1.4 m long string that is fixed at the other end. Initially at rest, he mass is made to rotate around the fixed end with an angular acceleration of 3.31 rad/s. What centripetal force must act on the mass after 8 s so that it continues to move in a circular path

Respuesta :

At time [tex]t[/tex] seconds, the mass has angular speed

[tex]\omega = \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t[/tex]

and hence linear speed

[tex]v = (1.4\,\mathrm m) \omega = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t[/tex]

After 8 s, its linear speed is

[tex]v = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) (8\,\mathrm s) = 37.072 \dfrac{\rm m}{\rm s} \approx 37 \dfrac{\rm m}{\rm s}[/tex]

and has centripetal acceleration with magnitude

[tex]a = \dfrac{v^2}{1.4\,\rm m} \approx 981.667\dfrac{\rm m}{\mathrm s^2} \approx 980 \dfrac{\rm m}{\mathrm s^2}[/tex]

To maintain this linear speed, by Newton's second law the required centripetal force should have magnitude

[tex]F = (0.34\,\mathrm{kg}) a \approx 333.767\,\mathrm N \approx \boxed{330 \,\mathrm N}[/tex]

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