ON07 P4 Q1 Elastic String Rotation | A2 Circular Motion | CAIE A Level 9702 Physics

9702/4/O/N/07: An elastic cord has an unextended length of 13.0 cm. One end of the cord is attached to a fixed point C. A small mass of weight 5.0 N is hung from the free end of the cord. The cord extends to a length of 14.8 cm, as shown in Fig. 1.1. #A2verticalCircleP4 #Lv3 #ASdeformationP2

The cord and mass are now made to rotate at constant angular speed ω in a vertical plane about point C. When the cord is vertical and above C, its length is the unextended length of 13.0 cm, as shown in Fig. 1.2.
(i) Show that the angular speed ω of the cord and mass is 8.7 rad s–1
(ii) The cord and mass rotate so that the cord is vertically below C, as shown in Fig. 1.3.Calculate the length L of the cord, assuming it obeys Hooke’s law.

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[UzairKhanNS]

That was helpful and easy to follow . You went through it step by step, gave the formulas needed ( I had totally forgotten Hooke's Law ) and you're personality made it less boring. From a new A2 student: Thanks!

[ibrahimshakir]

you forgot to square the angular speed in the last

[JudeWeraduwage]

Thank you for tension=0, very helpful!!!!

[raaziaabdullah]

Thanks for explaining !! You’re amazing 😭💪🏼✨

[asifjahanshuvro]

Hey, you forgot to square the angular speed.

[imboredd]

Hello miss, why doesnt the centripetal force remains at 5N according to when the mass is at the top? and so tension is 5+5= 10N when the mass is vertically below C?

[alkadas]

I didnt understand why t=0 can someone explain?
lol nvm I understood :)