A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

Solution 12E Step 1: Torque = F R Where, F - Force applied, R - radius of the sphere Or, torque = I Where, I - Moment of inertia, - angular acceleration So, F R = I 2 Provided, F = 40 N, R = 0.250 m, I = ½ MR (for solid uniform wheel) Where, M - mass of the wheel and R - radius of the wheel I = ½ × 9.20 kg × (0.250 m) = 4.60 kg × 0.0625 m = 0.2875 kg m 2 2 40 N × 0.250 m = 0.0029 kg m × Rearranging, = (40 N × 0.250 m) / 0.2875 kg m = 34.78 rad/s 2 Angular acceleration of the disc, = 34.78 rad/s 2 Step 2: 2 2 b) linear acceleration, a = R = 34.78 rad/s × 0.250 m = 8.695 m/s Applied force is, F = 40 N i Since the pull is acting towards the right of the horizontal direction, the applied force will be towards left. The force of gravitation will be acted on the wheel. It can be given as, F’ = mg = 9.20 kg × 9.8 m/s = 90.16 N It will be acting upward (y - direction) Therefore, F = 90.16 N j Magnitude of the resultant force, 2 2 2 2 F = F + F =40 90.16 = 1600 + 8128.82 = 9728.82 F = 98.63 N F = 40 N i + 90.16 N j tan = 90.16 N / 40 N = - 2.254 = tan ( - 2.254 ) = -66° It is in second quadrant. So, in order to get the direction, we should add 180 with this angle, Direction of the force vector, 180° - 66° = 114°