A drum of radius R and mass M rolls down with | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) When a body rolls down an inclined plane without slipping,the force of static friction acts on it. This frictional force provides the necessary to

Vedclass · Accepted Answer

Converts translational energy to rotational energy

Vedclass · Answer

(D) When a body rolls down an inclined plane without slipping,the force of static friction acts on it. This frictional force provides the necessary torque to initiate and maintain the rotational motion of the drum. As the drum moves down,the gravitational potential energy is converted into both translational kinetic energy and rotational kinetic energy. The frictional force acts to convert a portion of the translational energy into rotational energy,ensuring the condition of pure rolling (no slipping) is maintained. Therefore,the frictional force converts translational energy into rotational energy.

$A$ drum of radius $R$ and mass $M$ rolls down without slipping along an inclined plane of angle $\theta$. The frictional force

Similar Questions

Consider the circle $x^2+y^2-4x-2y+c=0$ whose centre is $A(2,1)$. If the point $P(10,7)$ is such that the line segment $PA$ meets the circle at $Q$ with $PQ=5$,then $c$ is equal to

Two solid spheres $A$ and $B$ made of the same material have radii $r_A$ and $r_B$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature of $A$ and $B$ is :

If $C$ is the midpoint of $AB$ and $P$ is any point outside $AB$,then:

Global agreement in specific control strategies to reduce the release of ozone depleting substances,was adopted by

$\frac{1}{2}-\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}-\frac{1}{4 \cdot 2^4}+\ldots$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module