ma = T mg – Use a = Example 1.36: (2012 Question 7c, 33%) = = 49 m s-2 Consider the motion to be vertical, at constant 2.0 x 49 = T 2.0 x 10 – speed. T = 118 N (ANS) Example 1.38: (2015 Question 3b, 50%) At the top of the path, there are two forces acting on the sphere, its weight, mg, and the tension from the string T. = ma Use FR = ma ΣF Becomes T + mg = = N mg – T = - mg. = N 2 × 10 – At the bottom of the circle, the situation looks N = 18 + 20 like. N = 38 N (ANS) Example 1.39: (NSW 2000 Question 2) For all circular motion the net force is radially inwards and the velocity is tangential. A (ANS) Question 4, 83%) Example 1.40: (1971 This requires you to express the KE in terms of the momentum and the mass. Use the relationship KE = K = (ANS) Example 1.41: (1971 Question 5, 76%) = ma ΣF The total energy, T, is constant. Becomes T - mg = Initially U = 0, therefore T = + mg. T = KE + PE Since is constant, the tension at the bottom is T = KE + 0 2mg larger than the force at the top. KE = T the tension in the string is greater at the A (ANS) bottom of the circular path. Question 6, 45%) Example 1.42: (1971 Example 1.37: (2013 Question 5b, 47%) To find the velocity at the highest point, you At the point S there are two forces acting on need to also know the PE at that point. This is the mass. The weight is acting down, and the not given. tension force is up. At the top, the vertical component of the Since the net force is up, velocity will be zero. The horizontal component = ma ΣF
Example 1.36: (2012 Question 7c, 33%) Consider the motion to be vertical, at constant speed. At the top of the path, there are two forces acting on the sphere, its weight, mg, and the tension from the string T. ΣF = ma Becomes T + mg = T = - mg. At the bottom of the circle, the situation looks like. ΣF = ma Becomes T - mg = T = + mg. Since is constant, the tension at the bottom is 2mg larger than the force at the top. the tension in the string is greater at the bottom of the circular path. Example 1.37: (2013 Question 5b, 47%) At the point S there are two forces acting on the mass. The weight is acting down, and the tension force is up. Since the net force is up, ΣF = ma ma = T – mg Use a = = = 49 m s-2 2.0 x 49 = T – 2.0 x 10 T = 118 N (ANS) Example 1.38: (2015 Question 3b, 50%) Use FR = ma = N – mg = N – 2 × 10 N = 18 + 20 N = 38 N (ANS) Example 1.39: (NSW 2000 Question 2) For all circular motion the net force is radially inwards and the velocity is tangential. A (ANS) Example 1.40: (1971 Question 4, 83%) This requires you to express the KE in terms of the momentum and the mass. Use the relationship KE = K = (ANS) Example 1.41: (1971 Question 5, 76%) The total energy, T, is constant. Initially U = 0, therefore T = KE + PE T = KE + 0 KE = T A (ANS) Example 1.42: (1971 Question 6, 45%) To find the velocity at the highest point, you need to also know the PE at that point. This is not given. At the top, the vertical component of the velocity will be zero. The horizontal component
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