More practice questions

  1. Q4. A force F = 5i^\hat i N acts at the point r = (3i^\hat i + 4j^\hat j) m. Find the torque about the origin, and say whether it turns the body clockwise or anticlockwise.

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    Both vectors lie in the xy-plane, so only the z-component survives: τz = xFy − yFx = 3(0) − 4(5) = −20. So τ = −20k^\hat k N m, into the page: a clockwise turning effect. Check: the line of action (the line y = 4 m) is 4 m from the origin, and 4 × 5 = 20 N m.

  2. Q5. A ball of mass 0.5 kg is thrown horizontally at 4 m s from the origin. Taking g = 10 m s, find its angular momentum about the origin 1 s later, and the torque of gravity about the origin at that moment.

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    After 1 s, with x along the throw and y upward, it is at (4, −5) m with velocity (4, −10) m s. The z-component of r × mv is 0.5[4(−10) − (−5)(4)] = 0.5(−40 + 20) = −10, so l=−10 k^{\mathbf{l} = -10\,\hat k} kg m s. The weight, 0.5 × 10 = 5 N along −j^-\hat j, acts on a moment arm of 4 m, so τ=−20 k^{\boldsymbol{\tau} = -20\,\hat k} N m. Check: in general lz = −10t, whose rate of change at t = 1 s is −20, equal to the torque.

  3. Q6. In an ideal central gravitational field, a comet orbits a fixed star on a long ellipse. At its nearest point it is 1.0 × 10 m from the star and moves at 50 km s. How fast does it move at its farthest point, 5.0 × 10 m from the Sun?

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    Gravity points at the fixed star, so its torque about the star is zero and the comet’s angular momentum about that point is constant. At both points the velocity is at right angles to the line to the Sun, so mv1r1 = mv2r2, and v2 = 50 × (1.0 × 10)/(5.0 × 10) = 1.0 km s.