A force of 49mathrm~N$49\mathrm{~N}$ acts tangentially at the highest point of a sphere (solid) of mass 20mathrm~kg$20\mathrm{~kg}$, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
A schematic of a solid sphere of mass m resting on a horizontal plane with a force F pointing horizontally to the right at the highest point.
A.3.5mathrm~m/s^2$3.5\mathrm{~m/s}^2$
B.0.35mathrm~m/s^2$0.35\mathrm{~m/s}^2$
C.2.5mathrm~m/s^2$2.5\mathrm{~m/s}^2$
D.0.25mathrm~m/s^2$0.25\mathrm{~m/s}^2$
Solution & Explanation
### Related Formula
Torque equation about the instantaneous center of zero velocity (bottom contact point P$P$):
tau_P = I_P alpha$\tau_P = I_P \alpha$
For a solid sphere, the moment of inertia about the center is I_c = frac25MR^2$I_c = \frac{2}{5}MR^2$. By the parallel axis theorem:
I_P = I_c + MR^2 = frac75MR^2$I_P = I_c + MR^2 = \frac{7}{5}MR^2$
### Core Logic
Since the sphere rolls without slipping, we can conveniently write the torque equation about the lowest point of contact P$P$ because static friction passes through this point and exerts zero torque.
- Distance from point P$P$ to the top highest point is 2R$2R$.
- Tangential force F = 49mathrm~N$F = 49\mathrm{~N}$.
- Mass of solid sphere, M = 20mathrm~kg$M = 20\mathrm{~kg}$.
tau_P = F times 2R$\tau_P = F \times 2R$
Substitute tau_P$\tau_P$ and I_P$I_P$ into the torque equation:
F times 2R = left(frac75MR^2right) alpha$F \times 2R = \left(\frac{7}{5}MR^2\right) \alpha$
### Step 1: Solving for Linear Acceleration
For pure rolling, the acceleration of the center of mass a$a$ is related to angular acceleration alpha$\alpha$ by a = Ralpha$a = R\alpha$:
2F R = frac75MR^2 left(fracaRright)$2F R = \frac{7}{5}MR^2 \left(\frac{a}{R}\right)$2F = frac75 M a implies a = frac10F7M$2F = \frac{7}{5} M a \implies a = \frac{10F}{7M}$
Substitute the numerical values (F = 49mathrm~N$F = 49\mathrm{~N}$ and M = 20mathrm~kg$M = 20\mathrm{~kg}$):
a = frac10 times 497 times 20 = frac490140 = 3.5mathrm~m/s^2$a = \frac{10 \times 49}{7 \times 20} = \frac{490}{140} = 3.5\mathrm{~m/s}^2$
### Step 2: Analysis of Friction Force Direction
Let's write force equations to verify consistency:
F + f = M a$F + f = M a$49 + f = 20 times 3.5 = 70 implies f = 21mathrm~N$49 + f = 20 \times 3.5 = 70 \implies f = 21\mathrm{~N}$
Since f$f$ is positive, static friction acts in the forward direction. Rolling without slipping is fully maintained since the required static friction coefficient is well within realistic limits.
### Pattern Recognition
Calculating torque about the bottom contact point is a powerful shortcut for rolling-without-slipping questions! It completely bypasses having to guess or set up equations for the friction direction.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
A schematic of a solid sphere of mass m resting on a horizontal plane with a force F pointing horizontally to the right at the highest point.
Keywords:#solid sphere pure rolling acceleration#torque about contact point#JEE Main 2025 Morning Q10#friction in pure rolling#pure rolling#torque equation#solid sphere rolling
More Rotational Motion Previous-Year Questions
Q32025Moment of Inertia and Torque
A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10mathrm~kg$10\mathrm{~kg}$ and radius is 10mathrm~cm$10\mathrm{~cm}$ and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20mathrm~N$20\mathrm{~N}$ is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1mathrm~m$1\mathrm{~m}$, would be:
A wheel rotating with a pull force of 20 N acting on the cord.
A.20mathrm~rad/s$20\mathrm{~rad/s}$
B.30mathrm~rad/s$30\mathrm{~rad/s}$
C.10mathrm~rad/s$10\mathrm{~rad/s}$
D.0mathrm~rad/s$0\mathrm{~rad/s}$
Solution
### Related Formula
W_F = F cdot s$W_F = F \cdot s$K_R = frac12 I omega^2$K_R = \frac{1}{2} I \omega^2$I = M R^2 quad text(Moment of inertia of a ring/rim)$I = M R^2 \quad \text{(Moment of inertia of a ring/rim)}$
### Core Logic
The work done by the constant pulling force F = 20mathrm~N$F = 20\mathrm{~N}$ through distance s = 1mathrm~m$s = 1\mathrm{~m}$ is completely converted into the rotational kinetic energy of the wheel.
Work done by force:
W_F = F cdot s = 20 times 1 = 20mathrm~J$W_F = F \cdot s = 20 \times 1 = 20\mathrm{~J}$
Since the spokes are of negligible mass, all mass M = 10mathrm~kg$M = 10\mathrm{~kg}$ is distributed on the outer rim of radius R = 10mathrm~cm = 0.1mathrm~m$R = 10\mathrm{~cm} = 0.1\mathrm{~m}$. The wheel acts as a thin ring:
I = M R^2 = 10 times (0.1)^2 = 0.1mathrm~kgcdot m^2$I = M R^2 = 10 \times (0.1)^2 = 0.1\mathrm{~kg\cdot m^2}$
Using the work-energy theorem:
W_F = Delta K_R = frac12 I omega^2$W_F = \Delta K_R = \frac{1}{2} I \omega^2$20 = frac12 times 0.1 times omega^2$20 = \frac{1}{2} \times 0.1 \times \omega^2$40 = 0.1 omega^2 implies omega^2 = 400 implies omega = 20mathrm~rad/s$40 = 0.1 \omega^2 \implies \omega^2 = 400 \implies \omega = 20\mathrm{~rad/s}$
### Step 1: Final Conclusion
The angular velocity of the wheel after the cord is unwound by 1mathrm~m$1\mathrm{~m}$ is 20mathrm~rad/s$20\mathrm{~rad/s}$.
### Pattern Recognition
Work done in unwinding a string is F cdot s$F \cdot s$. Under pure rotation with zero friction, this is exactly frac12 I omega^2$\frac{1}{2} I \omega^2$. Always identify the mass distribution (here, rim with massless spokes acts as a thin cylinder/ring I = MR^2$I = MR^2$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
Class 11 Physics: Work, Energy and Power
Q132025Moment of Inertia and Torque
A square Lamina OABC of length 10mathrm~cm$10\mathrm{~cm}$ is pivoted at 'O'. Forces act on the Lamina as shown in the figure. If the Lamina remains stationary, then the magnitude of F$F$ is:
A square lamina OABC with multiple force vectors acting on its vertices, pivoted at O.
A.20mathrm~N$20\mathrm{~N}$
B. 0 (zero)
C.10mathrm~N$10\mathrm{~N}$
D.10sqrt2mathrm~N$10\sqrt{2}\mathrm{~N}$
Solution
### Related Formula
tau_O = F cdot r_perp$\tau_O = F \cdot r_{\perp}$sum tau_O = 0 quad text(for rotational equilibrium)$\sum \tau_O = 0 \quad \text{(for rotational equilibrium)}$
### Core Logic
Let the side length of the square lamina be l = 10mathrm~cm$l = 10\mathrm{~cm}$.
The lamina is pivoted at point O(0,0)$O(0,0)$ and remains stationary under rotational equilibrium. Therefore, the net torque about O$O$ must be zero.
Evaluating torque contributions about point O$O$:
- Forces acting directly at pivot O$O$ produce zero torque.
- Forces whose lines of action pass through O$O$ produce zero torque.
- The 10mathrm~N$10\mathrm{~N}$ force perpendicular to side OA$OA$ produces torque:
tau_1 = 10 times l quad text(Counter-Clockwise)$\tau_1 = 10 \times l \quad \text{(Counter-Clockwise)}$
- The unknown force F$F$ acting perpendicular to side OC$OC$ produces torque:
tau_2 = F times l quad text(Clockwise)$\tau_2 = F \times l \quad \text{(Clockwise)}$
Setting sum tau_O = 0$\sum \tau_O = 0$:
10 cdot l - F cdot l = 0 implies F = 10mathrm~N$10 \cdot l - F \cdot l = 0 \implies F = 10\mathrm{~N}$
### Step 1: Final Conclusion
The magnitude of the force $
### Step 1: Final Conclusion
The magnitude of the force $F is $ is $10\mathrm{~N}$.
### Pattern Recognition
In pivoted laminas, focus on the pivot and disregard any force vector whose line of action passes through the pivot. For symmetric placements, equate Clockwise torque = Counter-Clockwise torque directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
Q162025Moment of Inertia and Torque
Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is 'α'. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is:
A.alpha$\alpha$
B.alpha / 4$\alpha / 4$
C.alpha / 8$\alpha / 8$
D.alpha / 2$\alpha / 2$
Solution
### Related Formula
I = frac112 M L^2$I = \frac{1}{12} M L^2$
### Core Logic
Initially, the moment of inertia is:
alpha = fracM L^212$\alpha = \frac{M L^2}{12}$
When cut into two equal parts, each smaller rod has:
- Mass, m = fracM2$m = \frac{M}{2}$
- Length, l = fracL2$l = \frac{L}{2}$
When joined symmetrically as a cross, the target axis passes through their joint intersection perpendicular to their plane.
For each rod, this axis passes through its individual center of mass and is perpendicular to its length.
Thus, the total moment of inertia of the cross is the sum of the moments of inertia of the two rods:
I_textcross = I_1 + I_2 = 2 times left(frac112 m l^2right) = frac16 m l^2$I_{\text{cross}} = I_1 + I_2 = 2 \times \left(\frac{1}{12} m l^2\right) = \frac{1}{6} m l^2$
Substituting m = fracM2$m = \frac{M}{2}$ and l = fracL2$l = \frac{L}{2}$:
I_textcross = frac16 times left(fracM2right) times left(fracL2right)^2 = frac16 times fracM2 times fracL^24 = fracM L^248$I_{\text{cross}} = \frac{1}{6} \times \left(\frac{M}{2}\right) \times \left(\frac{L}{2}\right)^2 = \frac{1}{6} \times \frac{M}{2} \times \frac{L^2}{4} = \frac{M L^2}{48}$
Comparing with alpha$\alpha$:
I_textcross = frac14 left(fracM L^212right) = fracalpha4$I_{\text{cross}} = \frac{1}{4} \left(\frac{M L^2}{12}\right) = \frac{\alpha}{4}$
### Step 1: Final Conclusion
The moment of inertia of the cross is $
### Step 1: Final Conclusion
The moment of inertia of the cross is $\alpha / 4.
### Pattern Recognition
Since mass scales linearly ($.
### Pattern Recognition
Since mass scales linearly ($M \propto L), cutting a rod into $), cutting a rod into $n equal segments scales the length by $ equal segments scales the length by $1/n and mass by $ and mass by $1/n. The moment of inertia of each segment scales as $. The moment of inertia of each segment scales as $1/n^3. Reassembling $. Reassembling $n segments linearly sums their contributions, so the final moment of inertia scales as $ segments linearly sums their contributions, so the final moment of inertia scales as $n \times \frac{1}{n^3} = \frac{1}{n^2}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
Q22025Moment of Inertia
A rod of linear mass density lambda^prime$\lambda^{\prime}$ and length L$L$ is bent to form a ring of radius R$R$. Moment of inertia of ring about any of its diameter is:
### Related Formula
I_textdia = frac12 M R^2$I_{\text{dia}} = \frac{1}{2} M R^2$
where,
I_textdia$I_{\text{dia}}$ = moment of inertia of a ring about its diameter
M$M$ = total mass of the ring
R$R$ = radius of the ring
### Core Logic
Since the linear mass density is lambda^prime$\lambda^{\prime}$ (or lambda$\lambda$ as per the options), the total mass M$M$ of the rod of length L$L$ is:
M = lambda L$M = \lambda L$
When this rod is bent into a ring of radius R$R$, its circumference equals the length of the rod:
2pi R = L implies R = fracL2pi$2\pi R = L \implies R = \frac{L}{2\pi}$
Substituting M$M$ and R$R$ into the formula for the moment of inertia about the diameter:
I_textdia = frac12 M R^2 = frac12 (lambda L) left(fracL2piright)^2 = fraclambda L^38pi^2$I_{\text{dia}} = \frac{1}{2} M R^2 = \frac{1}{2} (\lambda L) \left(\frac{L}{2\pi}\right)^2 = \frac{\lambda L^3}{8\pi^2}$
### Pattern Recognition
Sees: "Rod of length L$L$ bent to form a ring" → R = fracL2pi$R = \frac{L}{2\pi}$.
Trap: Moment of inertia about the central axis perpendicular to the plane is MR^2$MR^2$, but about its diameter, it is half, i.e., frac12MR^2$\frac{1}{2}MR^2$.
Shortcut: I = frac12 (lambda L) left(fracL2piright)^2 = fraclambda L^38pi^2$I = \frac{1}{2} (\lambda L) \left(\frac{L}{2\pi}\right)^2 = \frac{\lambda L^3}{8\pi^2}$. ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
Q242025Angular Acceleration and Torque
A thin solid disk of 1mathrm~kg$1\mathrm{~kg}$ is rotating along its diameter axis at the speed of 1800mathrm~rpm$1800\mathrm{~rpm}$. By applying an external torque of 25pimathrm~Ncdot m$25\pi\mathrm{~N\cdot m}$ for 40mathrm~s$40\mathrm{~s}$, the speed increases to 2100mathrm~rpm$2100\mathrm{~rpm}$. The diameter of the disk is ________ mathrm~m$\mathrm{~m}$.
Numerical Answer.Answer: 40 to 40
Solution
### Related Formula
omega = 2pi fracN60$\omega = 2\pi \frac{N}{60}$omega_f = omega_i + alpha t$\omega_f = \omega_i + \alpha t$tau = I alpha$\tau = I \alpha$I_textdia = frac14 m R^2$I_{\text{dia}} = \frac{1}{4} m R^2$
where,
N$N$ = rotational speed in rpm
alpha$\alpha$ = angular acceleration
tau$\tau$ = torque applied
I_textdia$I_{\text{dia}}$ = moment of inertia of a solid disk about its diameter axis
### Core Logic
Given parameters:
- Mass, m = 1mathrm~kg$m = 1\mathrm{~kg}$
- Initial speed, N_i = 1800mathrm~rpm implies omega_i = frac1800 times 2pi60 = 60pimathrm~rad/s$N_i = 1800\mathrm{~rpm} \implies \omega_i = \frac{1800 \times 2\pi}{60} = 60\pi\mathrm{~rad/s}$
- Final speed, N_f = 2100mathrm~rpm implies omega_f = frac2100 times 2pi60 = 70pimathrm~rad/s$N_f = 2100\mathrm{~rpm} \implies \omega_f = \frac{2100 \times 2\pi}{60} = 70\pi\mathrm{~rad/s}$
- Time, t = 40mathrm~s$t = 40\mathrm{~s}$
- Torque, tau = 25pimathrm~Ncdot m$\tau = 25\pi\mathrm{~N\cdot m}$
First, calculate the angular acceleration alpha$\alpha$:
omega_f = omega_i + alpha t implies 70pi = 60pi + alpha (40)$\omega_f = \omega_i + \alpha t \implies 70\pi = 60\pi + \alpha (40)$alpha = frac10pi40 = fracpi4mathrm~rad/s^2$\alpha = \frac{10\pi}{40} = \frac{\pi}{4}\mathrm{~rad/s}^{2}$
Now relate torque to moment of inertia:
tau = I alpha implies 25pi = left( frac14 m R^2 right) left( fracpi4 right)$\tau = I \alpha \implies 25\pi = \left( \frac{1}{4} m R^2 \right) \left( \frac{\pi}{4} \right)$25pi = frac14 (1) R^2 fracpi4 implies 25pi = R^2 fracpi16$25\pi = \frac{1}{4} (1) R^2 \frac{\pi}{4} \implies 25\pi = R^2 \frac{\pi}{16}$R^2 = 400 implies R = 20mathrm~m$R^2 = 400 \implies R = 20\mathrm{~m}$
### Step 1: Compute Diameter
The question asks for the diameter of the disk (D$D$):
D = 2R = 2 times 20 = 40mathrm~m$D = 2R = 2 \times 20 = 40\mathrm{~m}$
### Pattern Recognition
Sees: Torque acting on a rotating disk increasing its speed.
Trap: The axis of rotation is the *diameter* axis, not the normal geometric center axis. This means the moment of inertia is I = frac14 m R^2$I = \frac{1}{4} m R^2$, not frac12 m R^2$\frac{1}{2} m R^2$! Check this detail carefully. ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
More Rotational Motion Questions — jee_main_2025_03_april_morning
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