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1. Hello I'm stuck on this problem, just wondering if anyone could help me, i've managed to do part a but i have no idea where to start on the other parts, i got 0.8rads^-2 for part a, if anyone could explain the others i would be thankful .
2. since the rope is passing over the wheel without slipping part B just needs you to divide the linear distance traveled by the lift in 5.0 secs by the circumference of the wheel.

part C wants you to do a vector addition of the centrepetal acceleration after 2.0 secs and the linear acceleration
3. (Original post by Joinedup)
since the rope is passing over the wheel without slipping part B just needs you to divide the linear distance traveled by the lift in 5.0 secs by the circumference of the wheel.

part C wants you to do a vector addition of the centrepetal acceleration after 2.0 secs and the linear acceleration
Hi thanks for the response but I'm still a bit stuck, how would you calculate the linear distance and how would you do a vector addition between them, and would you use suvat equations to find the linear acceleration?
4. (Original post by Scary)
Hi thanks for the response but I'm still a bit stuck, how would you calculate the linear distance and how would you do a vector addition between them, and would you use suvat equations to find the linear acceleration?
yes, I'd use suvat to find the linear distance for part B

for part C I'd find the linear speed at 2.0 seconds with suvat, this has to be the speed of a point (in fact every point) on the rim of the wheel and then get the centripetal acceleration from that. The linear acceleration is just the 0.4 m/s2 given in the question and you have to add these together at 90 degrees because the centripetal acceleration is normal to the edge of the wheel and the linear acceleration is tangential. you're only interested in the magnitude of the answer.
5. (Original post by Joinedup)
yes, I'd use suvat to find the linear distance for part B

for part C I'd find the linear speed at 2.0 seconds with suvat, this has to be the speed of a point (in fact every point) on the rim of the wheel and then get the centripetal acceleration from that. The linear acceleration is just the 0.4 m/s2 given in the question and you have to add these together at 90 degrees because the centripetal acceleration is normal to the edge of the wheel and the linear acceleration is tangential. you're only interested in the magnitude of the answer.
ah okay thank you. i understand now

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