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Question: A smooth conical shell with its axis aligned to the vertical and apex pointing downwards has its surface at an angle of 30 degrees to the horizontal. A ball of mass m = 1kg moves in circular motion around the inside of the conical shell at a height of h = 2m above the apex. The ball is attached by a taught, light, and inextensible string to a mass of M = 2kg directly below the apex of the cone. The string passes through a small hole in the apex of the cone. What speed must the ball have to remain at a constant height above the surface of the cone.Answer:Resolving the horizontal and vertical components of the force we find. mv²/r = Ncos(60) + Tcos(30) equation (1) and Nsin(60) = mg + Tsin(30) equation (2). As, m has a constant height and the string is inextensible, M must also have a constant height. Therefore, the tension in the string is equal to the weight of M => T = Mg =>T = 2g. Substituting T = 2g and m = 1 into equation (2). Nsin(60) = g + g = 2g. Dividing both sides by sin(60) gives. N = 4g/sqrt(3). Substituting the values of N and T into equation (1). mv²/r = 4gcos(60)/sqrt(3) + 2gcos(30) = mv²/r = 2g/sqrt(3) + gsqrt(3) equation (3) Using tan(x) = Opposite side/adjacent side to find the radius of rotation r. r/h = tan(60)=> r = 2tan(60) = 2sqrt(3)Substituting r and m into equation (3) and multiplying both sides by r gives v² = 4g + 6g => v = sqrt(10g)