Board and spool

7678 /www/nrich/html/content/id/7678/ 1 0000-00-00T00:00:00 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> A man is holding one end of a board and the another end is on a spool with the outer radius $R$ and the inner radius $r$. A board does not slip on the spool and the spool does not slip on the ground. The man starts to move with the board with a speed $u$. <mdo:image alt="" src="spool1.bmp" style="width: 700px; height: 317px;"></mdo:image> 1) How long does it take for the man to reach the spool? 2) Find the distance which must be traveled by the man to reach the spool. 3) Find the distance traveled by the man if $r = R$. 4) Calculate the time and the distance if $l = 298$cm, $R = 101$ cm, $r = 86$ cm, $u = 1$m/s. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><mdo:image alt="" src="spool.bmp" style="width: 400px; height: 181px;"></mdo:image> 1) Suppose that the spool is rotating with an angular speed $\omega$ about a point A. The spool is a rigid body, this means that the angular speed is the same for all points. Write equations for points B and C: $$\omega_C = \frac{v_C}{R}, \omega_B = \frac{v_B}{R + r}$$ but $\omega_B = \omega_C = \omega$ and $v_B = u$ because the board does not slip on the spool. Thus, $v_C = \frac{R}{R+r}u$. This means that the speed at which the man is approching the spool is $v = u - v_C = u - \frac{R}{R+r}u = \frac{r}{R+r}u$. This means that the time needed for the man to reach the spool is $$t = \frac{l}{v} = \frac{l(R+r)}{ru}$$ 2) The man will travel $s = ut = \frac{l(R+r)}{r}$. 3) If $r = R$ then $s = 2l$. 4) Plug numbers to the equations but do not forget to change units, $t = 6.48$s and $s = 6.48$m. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Suppose that a cylinder is rotating about a point A with an angular speed $\omega$ and use the fact that an angular speed is the same for all points of the rigid body. There is another way of solving this problem. Look at the motion of spool as two seperate motions: the spool moves horizontally as its all mass is at point B and rotates about the point B. The angular speed is the rate of rotation $\omega = \frac{\delta \Theta}{\delta t}$. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <mdo:image alt="" src="spool.bmp" style="width: 400px; height: 181px;"></mdo:image> 1) Suppose that the spool is rotating with an angular speed $\omega$ about a point A. The spool is a rigid body, this means that the angular speed is the same for all points. Write equations for points B and C: $$\omega_C = \frac{v_C}{R}, \omega_B = \frac{v_B}{R + r}$$ but $\omega_B = \omega_C = \omega$ and $v_B = u$ because the board does not slip on the spool. Thus, $v_C = \frac{R}{R+r}u$. This means that the speed at which the man is approching the spool is $v = u - v_C = u - \frac{R}{R+r}u = \frac{r}{R+r}u$. This means that the time needed for the man to reach the spool is $$t = \frac{l}{v} = \frac{l(R+r)}{ru}$$ 2) The man will travel $s = ut = \frac{l(R+r)}{r}$. 3) If $r = R$ then $s = 2l$. 4) Plug numbers to the equations but do not forget to change units, $t = 6.48$s and $s = 6.48$m. </mdoxml> 2 4 0 0 1 1 1 Board and spool