Mechanics 3 Edexcel Circular Motion - help needed

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DanKeitley
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#1
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#1
Hi guys,

Really struggling with questions involving horizontal motion below the centre of a hollow sphere. Was wondering if someone could help me out by explaining how to solve the question attached. Also, if it is possible to give me a run down of the generic method for solving these types of questions, it would be much appreciated.

Thanks,
Dan
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user2020user
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(Original post by DanKeitley)
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For this question at the very least, you really want to draw a clear diagram of what's going on. Afterwards, you really want to isolate the normal reaction from the smooth and hollow sphere by resolving in a direction that is perpendicular to the direction of the weight of the mass m. To resolve in that direction, you'll need to find the value of the angle between the radius of the sphere, 2a, and the radius of the circular motion, \frac{3a}{2}. Basic trigonometry does that pretty quickly. Finally, you want to resolve in a direction that incorporates the weight of the mass m and the normal reaction which you should have found.
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brianeverit
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(Original post by DanKeitley)
Hi guys,

Really struggling with questions involving horizontal motion below the centre of a hollow sphere. Was wondering if someone could help me out by explaining how to solve the question attached. Also, if it is possible to give me a run down of the generic method for solving these types of questions, it would be much appreciated.

Thanks,
Dan
Resolve vertically to obtain the normal reaction in terms of the mass, then resolve horizontally and apply the condition for circular motion.
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DanKeitley
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(Original post by Khallil)
For this question at the very least, you really want to draw a clear diagram of what's going on. Afterwards, you really want to isolate the normal reaction from the smooth and hollow sphere by resolving in a direction that is perpendicular to the direction of the weight of the mass m. To resolve in that direction, you'll need to find the value of the angle between the radius of the sphere, 2a, and the radius of the circular motion, \frac{3a}{2}. Basic trigonometry does that pretty quickly. Finally, you want to resolve in a direction that incorporates the weight of the mass m and the normal reaction which you should have found.

(Original post by brianeverit)
Resolve vertically to obtain the normal reaction in terms of the mass, then resolve horizontally and apply the condition for circular motion.

Thanks for the help. For some reason, when trying to work out the angle, I forgot the distance from the particle to the centre of the circle was the radius, 2a. Haha :ashamed2::banghead::facepalm:
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