A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 rpm to 1500 rpm in an interval of 5 seconds. i. What is the linear velocity at the beginning and end of the interval? (5 marks) ii.What are the normal and tangential components of the acceleration of the midpoint of the bar after 5 seconds (5 marks) iii.The torque which must be applied to the drum at starting (5 marks) iv. The power at the end of the acceleration period (5 marks)

i. To find the linear velocity at the beginning and end of the interval, we first need to convert the rotational speeds from rpm to rad/s.<br /><br />At the beginning:<br />ω1 = 1200 rpm * (2π rad/60 s) = 125.66 rad/s<br /><br />At the end:<br />ω2 = 1500 rpm * (2π rad/60 s) = 157.08 rad/s<br /><br />Now, we can calculate the linear velocity using the formula v = rω, where r is the radius of the bar.<br /><br />At the beginning:<br />v1 = 1.5 m * 125.66 rad/s = 188.49 m/s<br /><br />At the end:<br />v2 = 1.5 m * 157.08 rad/s = 235.62 m/s<br /><br />Therefore, the linear velocity at the beginning of the interval is 188.49 m/s, and at the end is 235.62 m/s.<br /><br />ii. To find the normal and tangential components of the acceleration of the midpoint of the bar after 5 seconds, we first need to calculate the angular acceleration (α) using the formula α = (ω2 - ω1) / t, where t is the time interval.<br /><br />α = (157.08 rad/s - 125.66 rad/s) / 5 s = 6.048 rad/s²<br /><br />Now, we can calculate the angular velocity at the midpoint after 5 seconds using the formula ω = ω1 + αt.<br /><br />ω = 125.66 rad/s + 6.048 rad/s² * 5 s = 137.33 rad/s<br /><br />The normal component of the acceleration (an) can be calculated using the formula an = rω², where r is the radius of the bar.<br /><br />an = 1.5 m * (137.33 rad/s)² = 282.5 m/s²<br /><br />The tangential component of the acceleration (at) can be calculated using the formula at = rα.<br /><br />at = 1.5 m * 6.048 rad/s² = 9.072 m/s²<br /><br />Therefore, the normal component of the acceleration of the midpoint of the bar after 5 seconds is 282.5 m/s², and the tangential component is 9.072 m/s².<br /><br />iii. To find the torque which must be applied to the drum at starting, we can use the formula τ = Iα, where I is the moment of inertia of the bar.<br /><br />For a thin rod rotating about one end, the moment of inertia is given by I = (1/3)ml², where m is the mass of the rod and l is its length.<br /><br />Since the mass of the rod is not given, we cannot calculate the exact value of the torque. However, we can express it in terms of the mass of the rod.<br /><br />τ = (1/3)ml² * 6.048 rad/s²<br /><br />Therefore, the torque which must be applied to the drum at starting is (1/3)ml² * 6.048 rad/s².<br /><br />iv. To find the power at the end of the acceleration period, we can use the formula P = τω, where τ is the torque and ω is the angular velocity.<br /><br />We already know the angular velocity at the end of the acceleration period is 137.33 rad/s. We can use the same expression for torque as in part iii.<br /><br />P = (1/3)ml² * 6.048 rad/s² * 137.33 rad/s<br /><br />Therefore, the power at the end of the acceleration period is (1/3)ml² * 6.048 rad/s² * 137.33 rad/s.