Recent content by Thanuka

thank you for your help I appreaciate it

i think i got now Rcos theta= 6N as centripetal force is 6N Rsin theta = 3N as mass of the particle is 0.3kg therefore: tan theta = 1/2 theta = 26 deg and 34 mintues then from the triangle cos theta =20/radius we know theta therefore we can find cos theta cos theta = 0.8954 0.8954 =...

I think we have to resolve the normal reaction force exert by the bowl on the particle like "Rcos theta" then it is equal to the 6N which is centripetal force I don't know ,,,is it right??

ohh sorry its not the centripetal acceleration its the centriprtal force that goes through the center of the bowl

I think the centripetal force we found should be normal to the inner surface of the bowl as centripetal acceleration always goes through the center of the circle ,,,i got a triangle which one side is 20cm and its hypotenus is the centripetal force which is 6N After that i got stucked..

ohhh yes if it is smooth centripetal force is the only componant of the reaction force as there is no friction But how can i find the radius of the bowl??

Can anyone please answer this question 1.A glass marble of mass m is moving int a horizontal circle round the inside surface of a smooth hemispherical bowl of radius r. The centre of the circle is at a distance 1/2r below the centre of the bowl. (i) Find the magnitude of the reaction between...

For the above problem i got the answer. But i got another question I really apprecaite if anyone can give an answer for this question This is the question A particle of mass 0.3kg moves with an angular velocity of 10rad per second in a horizontal circle 20cm inside a smooth hemispherical...

A bucket of water is swung in a vertical circle of radius r in such a way that the bucket is upside down when it is at the top of the circle. What is the minimum speed that the bucket may have at this point if the water is to remain in it?