A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an

Q: Why do the two $1\text{ N}$ forces cancel out in the linear equation but not in the torque equation?

In the linear equation, we only care about the direction of the forces. Since one points up the incline and the other points down, their vector sum is zero. However, torque depends on *where* the force is applied. Because they are applied on opposite sides of the center of mass, they both create a twist in the same counter-clockwise direction, so their torques add up instead of canceling!

Q: How do we know the direction of static friction?

We usually assume a direction (e.g., up the incline) and let the math decide. If our calculated $f$ turns out positive, our assumption was correct. Here, friction must act up the incline to provide the necessary clockwise torque for the sphere to roll down without slipping.

Q: Why is the torque of the top force counter-clockwise?

Imagine the sphere is a steering wheel. If you push the top of the wheel to the left (up the incline), the wheel will turn counter-clockwise. Mathematically, $\vec{r} \times \vec{F}$ gives a vector pointing out of the page, which corresponds to a counter-clockwise rotation.

Q: Can we solve this by taking torque about the point of contact?

Absolutely! Taking torque about the instantaneous axis of rotation (the contact point) eliminates friction from the equation. You just need to remember to use the parallel axis theorem for the moment of inertia ($I_{\text{contact}} = \frac{7}{5} M R^2$) and calculate the distances of all forces from the contact point.

Q: What does 'rolling without slipping' physically mean here?

It means the point of the sphere touching the incline has zero velocity relative to the incline. This perfectly synchronizes the linear motion of the center of mass with the rotational motion of the sphere, giving us the golden constraint: $a = \alpha R$.

Question

A solid sphere of mass $1	ext{ kg}$ and radius $1	ext{ m}$ rolls without slipping on a fixed inclined plane with an angle of inclination $	heta = 30^\circ$ from the horizontal. Two forces of magnitude $1	ext{ N}$ each, parallel to the incline, act on the sphere, both at distance $r = 0.5	ext{ m}$ from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is______$	ext{ms}^{-2}$. (Take $g = 10	ext{ ms}^{-2}$ .)

Sigma Percentile · Accepted Answer

### The Setup: A Sphere Under Dual Influence

Imagine a solid sphere resting on a $30^\circ$ inclined plane. If we just let it go, gravity would pull it down, and static friction would cause it to roll smoothly. But this problem introduces a fascinating twist: two external forces, each of $1	ext{ N}$, are applied parallel to the incline.

The top force is applied at a distance $r = 0.5	ext{ m}$ above the center and points **up** the incline. The bottom force is applied at the same distance $r$ below the center but points **down** the incline. Our mission is to find the linear acceleration $a$ of the sphere as it rolls down without slipping.

### The Master Equation for Translation

Let's start by analyzing the forces acting along the incline. We define the downward direction along the incline as positive.

Gravity pulls the center of mass downward with a component of $mg \sin	heta$. The bottom $1	ext{ N}$ force also pushes downward, while the top $1	ext{ N}$ force and static friction $f$ oppose this motion by pointing upward.

Applying Newton's Second Law for translation ($F_{	ext{net}} = ma$):

$$ mg \sin 30^\circ + F_{	ext{bottom}} - F_{	ext{top}} - f = ma $$

Substituting the given values ($m = 1	ext{ kg}$, $g = 10	ext{ m/s}^2$, $\sin 30^\circ = 0.5$):

$$ 1(10)(0.5) + 1 - 1 - f = 1 \cdot a $$

Notice something beautiful? The two $1	ext{ N}$ forces perfectly cancel each other out in the linear direction! The equation simplifies elegantly to:

$$ 5 - f = a $$

### The Master Equation for Rotation

Now, we must shift our focus to the torques causing the sphere to rotate about its center of mass. Since the sphere is rolling down the incline, its angular acceleration $\alpha$ must be **clockwise**. We will take clockwise as the positive direction for torque.

Static friction $f$ acts at the bottom edge (distance $R = 1	ext{ m}$ from the center) and points up the incline. This creates a **clockwise** torque of $f \cdot R$.

What about our two applied forces? 
- The top force points up the incline. If you push the top of a wheel to the left, it turns counter-clockwise. Torque = $1 \cdot r$ (counter-clockwise).
- The bottom force points down the incline. If you push the bottom of a wheel to the right, it also turns counter-clockwise. Torque = $1 \cdot r$ (counter-clockwise).

Both applied forces are actively fighting the clockwise rotation! Applying Newton's Second Law for rotation ($	au_{	ext{net}} = I\alpha$):

$$ f \cdot R - F_{	ext{top}} \cdot r - F_{	ext{bottom}} \cdot r = \frac{2}{5} M R^2 \alpha $$

Substituting the values ($R = 1	ext{ m}$, $r = 0.5	ext{ m}$):

$$ f(1) - 1(0.5) - 1(0.5) = 0.4(1)(1)^2 \alpha $$

$$ f - 1 = 0.4 \alpha $$

### The Grand Synthesis

Because the sphere rolls without slipping, its linear and angular accelerations are perfectly synchronized by the golden constraint:

$$ a = \alpha R $$

Since $R = 1	ext{ m}$, we simply have $\alpha = a$. Substituting this into our torque equation gives:

$$ f - 1 = 0.4 a $$

We now have a coupled system of two equations:
1. $f = 5 - a$
2. $f - 1 = 0.4 a$

Let's substitute the first equation into the second to eliminate friction:

$$ (5 - a) - 1 = 0.4 a $$

$$ 4 - a = 0.4 a $$

### Final Calculation

Bringing all the $a$ terms to one side:

$$ 4 = 1.4 a $$

$$ a = \frac{4}{1.4} = \frac{40}{14} = \frac{20}{7} $$

Calculating the decimal value:

$$ a \approx 2.86 	ext{ m/s}^2 $$

And there we have it! The two applied forces didn't change the net linear force, but by creating a counter-clockwise torque, they forced the static friction to increase, which ultimately slowed down the sphere's acceleration.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Translational Force Balance$

$Simplifying Linear Equation$

$Rotational Torque Balance$

$Analyzing Torque Directions$

$Newton’s Second Law for Rotation$

$Rolling Without Slipping Constraint$

$Solving the Coupled System$

$Final Acceleration Calculation$

$Conceptual Reflection$

The Sigma Insight: Rolling Motion

The Setup

The Master Equation for Translation

The Master Equation for Rotation

The Grand Synthesis

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Translational Force Balance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying Linear Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rotational Torque Balance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Torque Directions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Newton’s Second Law for Rotation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rolling Without Slipping Constraint

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving the Coupled System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Acceleration Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conceptual Reflection