A 3.00 kgstone is tied to a thin, light wire wrapped around the outer edge of th

Question

A 3.00 kgstone is tied to a thin, light wire wrapped around the outer edge of the uniform 10.0kgcylindrical pulley shown in the figure below (Figure 1) . The inner diameter of the pulley is 60.0 cm , while the outer diameter is 1.00 m. The system is released from rest, and there is no friction at the axle of the pulley. A) Find the acceleration of the stone. B) Find the tension in the wire. C) Find the angular acceleration of the pulley. A 3.00 kgstone is tied to a thin, light wire wrapped around the outer edge of the uniform 10.0kgcylindrical pulley shown in the figure below (Figure 1) . The inner diameter of the pulley is 60.0 cm , while the outer diameter is 1.00 m. The system is released from rest, and there is no friction at the axle of the pulley. A) Find the acceleration of the stone. B) Find the tension in the wire. C) Find the angular acceleration of the pulley. A 3.00 kgstone is tied to a thin, light wire wrapped around the outer edge of the uniform 10.0kgcylindrical pulley shown in the figure below (Figure 1) . The inner diameter of the pulley is 60.0 cm , while the outer diameter is 1.00 m. The system is released from rest, and there is no friction at the axle of the pulley. A) Find the acceleration of the stone. B) Find the tension in the wire. C) Find the angular acceleration of the pulley. Find the acceleration of the stone. B) Find the tension in the wire. C) Find the angular acceleration of the pulley. Find the tension in the wire. C) Find the angular acceleration of the pulley. Find the angular acceleration of the pulley.

Explanation / Answer

mg-T=ma
3g-T=3a
Tr=Ia/R

=>R=1m

I = 1/2 m ( ri2 + ro2)=.5*10*(1+.6^2)=6.8kgm^2

T=6.8a

=>a=3g/(6.8+3)=3m/s^2

T=20.4N

=a/R=3rad/s^2

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