A uniform disc of mass `1 kg` and radius `20 cm` is rolling purely on a flat horizontal surface. Its centre `C` is moving with acceleration `a = 20 ms^(-2)` and velocity `v =4 m//s` at a certain instant. At this instant. points `A` and `B` are located on the disc as shown in the diagram, with `AC = BC = 10 cm`. What is the acceleration magnitude of point `A`?A. `10sqrt(17)ms^(-2)`B. `10sqrt(37)ms^(-2)`C. `10sqrt(35)ms^(-2)`D. `60sqrt(17)ms^(-2)`

A uniform disc of mass `1 kg` and radius `20 cm` is rolling purely on

1.	A uniform disc of mass `1 kg` and radius `20 cm` is rolling purely on a flat horizontal surface. Its centre `C` is moving with acceleration `a = 20 ms^(-2)` and velocity `v =4 m//s` at a certain instant. At this instant. points `A` and `B` are located on the disc as shown in the diagram, with `AC = BC = 10 cm`. What is the acceleration magnitude of point `A`?A. `10sqrt(17)ms^(-2)`B. `10sqrt(37)ms^(-2)`C. `10sqrt(35)ms^(-2)`D. `60sqrt(17)ms^(-2)`
Answer» Correct Answer - B `KE=1/2mv^(2)=1/2Iomega^(2)=3/4mv^(2)=12J` `veca_(A//C)=veca_(A)-veca_(C)` ` implies veca_(A)=veca_(A//C)+veca_(C)=omega^(2)rhati+alpharhatj+ahati=60hati+10hatj` `veca_(A)=sqrt(3600+10)=10sqrt(37)m//s^(2)` `veca_(B)=omega^(2)rhatj+alphar(-hati)+ahati=10hati+40hatj` `a_(B)=10sqrt(17)m//s^(2)`

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