torque angular velocity rev/min

Question

2.2 A point on the rim of a wheel with .a diameter of 500 mm has a velocity of 200 km/h. Calculate the following: 2.2.1 The revolutions per minute (rev/min) 2.2.2 The angular velocity in rad/s at which the wheel is turning 2.3 The engine of a vehicle develops 67 kW at a speed of 1 200 r/min. Calculate the torque developed.

Answer 1

2.2.1)

Given data

Diameter d = 500 mm

1 mm = (10^-3) m

d = 500×10^-3 m

d = 0.5 m

Velocity v = 200 km/h

1km/h = (5/18) m/s

v = 200 (5/18) m/s

v = 55.56 m/s

revolutions per minute (rev/min) = ?

The number of circumferences of wheel which fit inside the total distance is the number of times the wheel revolves( rev/min) in that time period.

The number of revolutions per minute = (speed) / (circumference of wheel)

The number of revolutions per minute = v / πd

= 55.56 / 0.5π

= 35.3857

= 35 rpm

rpm = The revolutions per minute

The number of revolutions of wheel per minute is 35

Answer 2

2.2.2)

Given data

Diameter d = 500 mm

1 mm = (10^-3) m

d = 500×10^-3 m

d = 0.5 m

radius r = d/2

r = 0.5/2

r = 0.25 m

linear speed v = 200 km/h

1km/h = (5/18) m/s

v = 200 (5/18) m/s

v = 55.56 m/s

Angular velocity  ω = ?

The angular velocity is defined as the rate of change of angular displacement.

The number of revolutions per minute = (speed) / (circumference of wheel)

The number of revolutions per minute (rpm) = v / πd

= 55.56 / 0.5π

= 35.3857 rpm

Angular velocity  ω= rpm x (2π)/60

ω = 35.387 x (2π)/60

ω = 35.387 x (2π)/60

ω = 3.70 rad/s

Solution :

The angular velocity of wheel is  3.7 radians per second

 

Answer 3

2.3)

Given data :

The vehicle engine developed power P =  67 kW = 67000 W

Speed N =1200 r/min.

Torque T = ?

Formula :

Power P = 2πNT/60

T = (60/2π)(P/N)

T = 9.5488(P/N)

Substitute Speed N =1200 rpm and P =  67000 W

T = 9.5488(67000/1200)

T = 533.14 N - m

(N.m = Newton meters )

The required torque T = 533.14 N - m