# The garden roller problem (trigonometry)

• Aug 16th 2005, 10:55 AM
Trigo
The garden roller problem (trigonometry)
The diagram below shows a simplified side view of a garden roller. The radius of the cylinder is 30 cm. The length of the handle is 130 cm. The roller is leaning against a wall with the handle at an angle of 25 degrees to the vertical the ground is horizontal.

A gardener takes the roller on to a lawn. Just as he begins to roll he notices a small scratch at the highest point of the cylinder.

(iv) Show that, when the cylinder has been pushed a distance x cm on the lawn, the height of the scratch above the ground is:

30{1+cos(6x/pi)}cm

Anyone know how to get there? :confused:
• Aug 16th 2005, 01:00 PM
rgep
What scratch?
• Aug 16th 2005, 01:22 PM
Trigo
I have now added the missing information. My apologies.
• Aug 16th 2005, 10:25 PM
rgep
If the roller has been pushed x cm along the ground, it has rotated by the x/(circumference) of a revolution: that is, by x/30 radians or 360x/(2.pi.30) degrees. So the height of the scratch above the level of the centre is now 30cos(60x/pi): draw a picture showing the roller rotated by 60x/pi. The centre is itself 30 cm above the ground.
• Aug 17th 2005, 05:31 PM
Trigo
Thank you :)