Question

A steel drill making 180 rpm is used to drill a hole in a block of steel. The mass of the steel block and the drill is 180 g. If the entire mechanical work is used up in producing heat and the rate of rise in temperature of the block and the drill is 0.5 °C/sec. Find (a) the rate of working of the drill in Watts and (b) torque required to drive the drill. Given: specific heat capacity of steel = 420 J/(kgK)

Answer 1

Answer:

a) 37.8 W

b) 2 Nm

Explanation:

180 g = 0.18 kg

We can also convert 180 revolution per minute to standard angular velocity unit knowing that each revolution is 2π and 1 minute equals to 60 seconds

180 rpm = 180*2π/60 = 18.85 rad/s

We can use the heat specific equation to find the rate of heat exchange of the steel drill and block:

$\dot{E} = mc\Delta \dot{t} = 0.18*420*0.5 = 37.8 J/s$

Since the entire mechanical work is used up in producing heat, we can conclude that the rate of work is also 37.8 J/s, or 37.8 W

The torque T required to drill can be calculated using the work equation

$E = T\theta$

$\dot{E} = T\dot{\theta} = T\omega$

$T = \frac{\dot{E}}{\omega} = \frac{37.8}{18.85} = 2 Nm$