A circular saw blade with radius 0.120 m starts from rest and turns in a vertical plane with a constant angular acceleration of 3.00 rev/s2. After the blade has turned through 155 rev, a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of 0.820 m to the floor. How far does the piece travel horizontally. from where it broke off the blade until it strikes the floor?
Solution 63P Step 1: We know that, the angular displacement, = t +½ t 0 2 Where, - Initial angular velocity 0 - angular acceleration t - time for angular displacement Provided, the blade starts moving from the rest. Therefore, = 0 rad/s 0 2 Therefore, we can write, = ½ t Provided, = 155 rev and = 3 rev/s 2 2 2 Substituting these values in the equation, we get, 155 rev = ½ × 3 rev/s × t Rearranging, t = ( 155 rev × 2) / 3 rev/s = 310 rev / 3 rev/s 2 2 2 t = 103.33 s Taking square root on both sides, t = 10.16 s Step 2: We can write, = + t 0 Where, - final angular velocity 0 initial angular velocity - Angular acceleration t - Time 2 2 But, we should convert the unit of angular acceleration from rev/s to rad/s = 3 rev/s 2 1 rev = 2 rad 2 2 Therefore, = 3 × 2 rad / s = 6 rad / s We know that, the initial angular velocity, = 0 rad0 Therefore, = 0 + 6 rad / s × 10.16 s = 191.41 rad/s Step 3: The velocity of the tip can be given by, v = r Where, - angular velocity r - radius of the circular saw blade Provided, r = 0.120 m Therefore, v = 191.41 rad/s × 0.120 m = 22.97 m/s