In Fig$\mathbf{\text{10-61}}$., four pulleys are connected by two belts. Pulley A (radius$\mathbf{\text{15\hspace{0.17em}cm}}$) is the drive pulley, and it rotates at.$\mathbf{\text{10\hspace{0.17em}rad/s}}$Pulley B (radius$\mathbf{\text{10\hspace{0.17em}cm}}$) is connected by belt $\mathbf{1}$to pulley A. Pulley B’ (radius$\mathbf{\text{5\hspace{0.17em}cm}}$) is concentric with pulley B and is rigidly attached to it. Pulley C (radius$\mathbf{\text{25\hspace{0.17em}cm}}$) is connected by belt $\mathbf{2}$to pulley B’. Calculate (a) the linear speed of a point on belt $\mathbf{1}$, (b) the angularspeed of pulley B, (c) the angular speed of pulley B’, (d) the linear speed of a point on belt$\mathbf{2}$, and (e) the angular speed of pulley C. (Hint: If the belt between two pulleys does not slip, the linear speeds at the rims of the two pulleys must be equal.)

1. Radius of pulley A is$r_a=15\text{\hspace{0.17em}cm}$
2. Radius of pulley B is$r_b=10\text{cm}$
3. Radius of pulley B’ is$r_{b^{\text{'}}}=5\text{\hspace{0.17em}cm}$
4. Radius of pulley C is$r_c=25\text{cm}$
5. Angular velocity of pulley A is${\omega }_a=10\text{\hspace{0.17em}rad/s}$
6. If the belt between two pulleys does not slip, the linear speed at the rims of the two pulleys must be equal.

The linear velocity is given as the rate of change of displacement with respect to time. The angular velocity is defined as the rate of change of angular displacement with respect to time. Find the linearvelocity of belt 1 from radius of pulley A and angular speed of pulley A. From linear speed of belt 1, find the remaining values.

The relation between linear and angular velocity is-

$\mathbf{v}\mathbf{=}\mathbf{r\omega }$

where, v is velocity, ris radius and$\mathbf{\omega }$is angular velocity.

Use the following formula to find linear speed,

$\begin{array}{c}{v}_a=r_a{\omega }_a\\ =0.15\text{\hspace{0.17em}m}\times 10\text{\hspace{0.17em}rad}/\text{s}\\ =1.5\text{m/s}\end{array}$

Hence,linear speed of a point on belt$1$is$1.5\text{\hspace{0.17em}m/s}$.

Linear speed of pulley B is also$1.5\text{\hspace{0.17em}m}/\text{s}$because the belt doesn’t slip.

So, angular speed is given as follows:

$\begin{array}{c}{\omega }_b=\frac{v_b}{r_b}\\ =\frac{1.5\text{\hspace{0.17em}m}/\text{s}}{0.10\text{\hspace{0.17em}m}}\\ =15\text{\hspace{0.17em}rad/s}\end{array}$

Hence,angular speed of pulley$\text{B}$is$15\text{\hspace{0.17em}rad/s}$.

Angular speed of pulley$\text{B'}$is also$15\text{rad/s}$because it is concentric to pulley A.

Hence, angular speed of pulley B’ is$15\text{rad/s}$ .

Linear speed is given as follows:

$\begin{array}{c}{v}_2=r_{b^{\text{'}}}\times {\omega }_{b^{\text{'}}}\\ =0.05\text{\hspace{0.17em}m}\times 15\text{\hspace{0.17em}rad}/\text{s}\\ =0.75\text{\hspace{0.17em}m/s}\end{array}$

Hence, linear speed of point on belt 2 is $0.75\text{m/s}$.

Angular speed is as follows:

$\begin{array}{c}{\omega }_c=\frac{v_2}{r_c}\\ =\frac{0.75\text{\hspace{0.17em}m}/\text{s}}{0.25\text{\hspace{0.17em}m}}\\ =3\text{rad/s}\end{array}$

Hence, angular speed of pulley C is $3\text{\hspace{0.17em}rad/s}$$3\text{\hspace{0.17em}rad/s}$.