Question

Figure 8-56 shows a solid, uniform cylinder of mass 7.00 kg and radius 0.450 m with a light string wrapped around it. A 3.00-N tension force is applied to the string, causing the cylinder to roll without slipping across a level surface as shown. (a) What is the angular acceleration of the cylinder? (b) Calculate the magnitude and direction of the frictional force that acts on the cylinder. Figure attached below

Answer 1

Answer:

a) The cylinder has an angular acceleration of 3.810 radians per square second, b) The frictional force has a magnitude of 9 newtons and has the same direction of tension force.

Explanation:

The external force exerted on string creates a tension force that tries to move the cylinder in translation, but it is opposed by the friction force between cylinder and ground that generates rolling on cylinder. The Free Body Motion on cylinder-string system is presented below as attachment. Given that cylinder is a rigid body in planar motion, two equations of equilibrium for translation and an equation of equilibrium for rotation are needed to represent the system, which are now described:

$\Sigma F_{x} = T + f = M\cdot R\cdot \alpha$

$\Sigma F_{y} = N - M\cdot g = 0$

$\Sigma M_{G} = (T-f)\cdot R = I_{G}\cdot \alpha$

Where:

$T$ - Tension, measured in newtons.

$f$ - Friction force, measured in newtons.

$M$ - Mass of the cylinder, measured in kilograms.

$R$ - Radius of the cylinder, measured in meters.

$\alpha$ - Angular acceleration, measured in radians per square second.

$N$ - Normal force from ground exerted on cylinder, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$I_{G}$ - Moment of inertia of the cylinder with respect to its center of mass, measured in kilogram-square meters.

The moment of inertia of the cylinder is:

$I_{G} = \frac{1}{2}\cdot M\cdot R^{2}$

a) The angular acceleration is determined by solving on first and third equation after eliminating  friction force:

$f = M\cdot R \cdot \alpha - T$

$(T-M\cdot R\cdot \alpha+T) \cdot R = I_{G}\cdot \alpha$

$2\cdot T\cdot R = (I_{G} + M\cdot R^{2})\cdot \alpha$

$\alpha = \frac{2\cdot T\cdot R}{I_{G}+M\cdot R^{2}}$

$\alpha = \frac{2\cdot T \cdot R}{\frac{1}{2}\cdot M\cdot R^{2}+M\cdot R^{2} }$

$\alpha = \frac{4\cdot T}{3\cdot M\cdot R}$

If $T = 3\,N$, $M = 7\,kg$ and $R = 0.45\,m$, then:

$\alpha = \frac{4\cdot (3\,N)}{(7\,kg)\cdot (0.45\,m)}$

$\alpha = 3.810\,\frac{rad}{s^{2}}$

The cylinder has an angular acceleration of 3.810 radians per square second.

b) The magnitude of the frictional force can be determined with the help of the following expression:

$f = M\cdot R \cdot \alpha - T$

Given that $T = 3\,N$, $M = 7\,kg$, $R = 0.45\,m$ and $\alpha = 3.810\,\frac{rad}{s^{2}}$, the magnitude of the friction force is:

$f = (7\,kg)\cdot (0.45\,m)\cdot \left(3.810\,\frac{rad}{s^{2}} \right)-3\,N$

$f = 9\,N$

The frictional force has a magnitude of 9 newtons and has the same direction of tension force.